Split (2)

train · 8.66k rows

db_name stringclasses 146 values	prompt stringlengths 310 4.81k
manufacturer	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # manufacturer ( Manufacturer_ID, Open_Year, Name, Num_of_Factories, Num_of_Shops ) # furniture ( Furniture_ID, Name, Num_of_Component, Market_Rate ) # furniture_manufacte ( Manufacturer_ID, Furniture_ID, Price_in_Dollar ) # # furniture_manufacte.Furniture_ID can be joined with furniture.Furniture_ID # furniture_manufacte.Manufacturer_ID can be joined with manufacturer.Manufacturer_ID # ### Question: # # Find the id and number of shops for the company that produces the most expensive furniture. # ### SQL: # # SELECT t1.manufacturer_id , t1.num_of_shops FROM manufacturer AS t1 JOIN furniture_manufacte AS t2 ON t1.manufacturer_id = t2.manufacturer_id ORDER BY t2.Price_in_Dollar DESC LIMIT 1 # ### End.
manufacturer	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # manufacturer ( Manufacturer_ID, Open_Year, Name, Num_of_Factories, Num_of_Shops ) # furniture ( Furniture_ID, Name, Num_of_Component, Market_Rate ) # furniture_manufacte ( Manufacturer_ID, Furniture_ID, Price_in_Dollar ) # # furniture_manufacte.Furniture_ID can be joined with furniture.Furniture_ID # furniture_manufacte.Manufacturer_ID can be joined with manufacturer.Manufacturer_ID # ### Question: # # Find the number of funiture types produced by each manufacturer as well as the company names. # ### SQL: # # SELECT count(*) , t1.name FROM manufacturer AS t1 JOIN furniture_manufacte AS t2 ON t1.manufacturer_id = t2.manufacturer_id GROUP BY t1.manufacturer_id # ### End.
manufacturer	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # manufacturer ( Manufacturer_ID, Open_Year, Name, Num_of_Factories, Num_of_Shops ) # furniture ( Furniture_ID, Name, Num_of_Component, Market_Rate ) # furniture_manufacte ( Manufacturer_ID, Furniture_ID, Price_in_Dollar ) # # furniture_manufacte.Furniture_ID can be joined with furniture.Furniture_ID # furniture_manufacte.Manufacturer_ID can be joined with manufacturer.Manufacturer_ID # ### Question: # # Give me the names and prices of furnitures which some companies are manufacturing. # ### SQL: # # SELECT t1.name , t2.price_in_dollar FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID # ### End.
manufacturer	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # manufacturer ( Manufacturer_ID, Open_Year, Name, Num_of_Factories, Num_of_Shops ) # furniture ( Furniture_ID, Name, Num_of_Component, Market_Rate ) # furniture_manufacte ( Manufacturer_ID, Furniture_ID, Price_in_Dollar ) # # furniture_manufacte.Furniture_ID can be joined with furniture.Furniture_ID # furniture_manufacte.Manufacturer_ID can be joined with manufacturer.Manufacturer_ID # ### Question: # # Find the market shares and names of furnitures which no any company is producing in our records. # ### SQL: # # SELECT Market_Rate , name FROM furniture WHERE Furniture_ID NOT IN (SELECT Furniture_ID FROM furniture_manufacte) # ### End.
manufacturer	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # manufacturer ( Manufacturer_ID, Open_Year, Name, Num_of_Factories, Num_of_Shops ) # furniture ( Furniture_ID, Name, Num_of_Component, Market_Rate ) # furniture_manufacte ( Manufacturer_ID, Furniture_ID, Price_in_Dollar ) # # furniture_manufacte.Furniture_ID can be joined with furniture.Furniture_ID # furniture_manufacte.Manufacturer_ID can be joined with manufacturer.Manufacturer_ID # ### Question: # # Find the name of the company that produces both furnitures with less than 6 components and furnitures with more than 10 components. # ### SQL: # # SELECT t3.name FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID JOIN manufacturer AS t3 ON t2.manufacturer_id = t3.manufacturer_id WHERE t1.num_of_component < 6 INTERSECT SELECT t3.name FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID JOIN manufacturer AS t3 ON t2.manufacturer_id = t3.manufacturer_id WHERE t1.num_of_component > 10 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Display the first name and department name for each employee. # ### SQL: # # SELECT T1.first_name , T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the first name and department name of all employees? # ### SQL: # # SELECT T1.first_name , T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # List the full name (first and last name), and salary for those employees who earn below 6000. # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE salary < 6000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full names and salaries for any employees earning less than 6000? # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE salary < 6000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Display the first name, and department number for all employees whose last name is "McEwen". # ### SQL: # # SELECT first_name , department_id FROM employees WHERE last_name = 'McEwen' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the first names and department numbers for employees with last name McEwen? # ### SQL: # # SELECT first_name , department_id FROM employees WHERE last_name = 'McEwen' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Return all the information for all employees without any department number. # ### SQL: # # SELECT * FROM employees WHERE department_id = "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are all the employees without a department number? # ### SQL: # # SELECT * FROM employees WHERE department_id = "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Display all the information about the department Marketing. # ### SQL: # # SELECT * FROM departments WHERE department_name = 'Marketing' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information about the Marketing department? # ### SQL: # # SELECT * FROM departments WHERE department_name = 'Marketing' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # when is the hire date for those employees whose first name does not containing the letter M? # ### SQL: # # SELECT hire_date FROM employees WHERE first_name NOT LIKE '%M%' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # On what dates were employees without the letter M in their first names hired? # ### SQL: # # SELECT hire_date FROM employees WHERE first_name NOT LIKE '%M%' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the full name (first and last), hire date, salary, and department number for those employees whose first name does not containing the letter M. # ### SQL: # # SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full name, hire date, salary, and department id for employees without the letter M in their first name? # ### SQL: # # SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the full name (first and last), hire date, salary, and department number for those employees whose first name does not containing the letter M and make the result set in ascending order by department number. # ### SQL: # # SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' ORDER BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full name, hire data, salary and department id for employees without the letter M in their first name, ordered by ascending department id? # ### SQL: # # SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' ORDER BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # what is the phone number of employees whose salary is in the range of 8000 and 12000? # ### SQL: # # SELECT phone_number FROM employees WHERE salary BETWEEN 8000 AND 12000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Return the phone numbers of employees with salaries between 8000 and 12000. # ### SQL: # # SELECT phone_number FROM employees WHERE salary BETWEEN 8000 AND 12000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display all the information of employees whose salary is in the range of 8000 and 12000 and commission is not null or department number does not equal to 40. # ### SQL: # # SELECT * FROM employees WHERE salary BETWEEN 8000 AND 12000 AND commission_pct != "null" OR department_id != 40 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Return all information about employees with salaries between 8000 and 12000 for which commission is not null or where their department id is not 40. # ### SQL: # # SELECT * FROM employees WHERE salary BETWEEN 8000 AND 12000 AND commission_pct != "null" OR department_id != 40 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full name (first and last name) and salary for all employees who does not have any value for commission? # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE commission_pct = "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Return the full names and salaries of employees with null commissions. # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE commission_pct = "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Display the first and last name, and salary for those employees whose first name is ending with the letter m. # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE first_name LIKE '%m' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Return the full names and salaries for employees with first names that end with the letter m. # ### SQL: # # SELECT first_name , last_name , salary FROM employees WHERE first_name LIKE '%m' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find job id and date of hire for those employees who was hired between November 5th, 2007 and July 5th, 2009. # ### SQL: # # SELECT job_id , hire_date FROM employees WHERE hire_date BETWEEN '2007-11-05' AND '2009-07-05' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the job ids and dates of hire for employees hired after November 5th, 2007 and before July 5th, 2009? # ### SQL: # # SELECT job_id , hire_date FROM employees WHERE hire_date BETWEEN '2007-11-05' AND '2009-07-05' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the first and last name for those employees who works either in department 70 or 90? # ### SQL: # # SELECT first_name , last_name FROM employees WHERE department_id = 70 OR department_id = 90 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full names of employees who with in department 70 or 90? # ### SQL: # # SELECT first_name , last_name FROM employees WHERE department_id = 70 OR department_id = 90 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the salary and manager number for those employees who is working under a manager. # ### SQL: # # SELECT salary , manager_id FROM employees WHERE manager_id != "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the salaries and manager ids for employees who have managers? # ### SQL: # # SELECT salary , manager_id FROM employees WHERE manager_id != "null" # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display all the details from Employees table for those employees who was hired before 2002-06-21. # ### SQL: # # SELECT * FROM employees WHERE hire_date < '2002-06-21' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information about employees hired before June 21, 2002? # ### SQL: # # SELECT * FROM employees WHERE hire_date < '2002-06-21' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display all the information for all employees who have the letters D or S in their first name and also arrange the result in descending order by salary. # ### SQL: # # SELECT * FROM employees WHERE first_name LIKE '%D%' OR first_name LIKE '%S%' ORDER BY salary DESC # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information about employees with D or S in their first name, ordered by salary descending? # ### SQL: # # SELECT * FROM employees WHERE first_name LIKE '%D%' OR first_name LIKE '%S%' ORDER BY salary DESC # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display those employees who joined after 7th September, 1987. # ### SQL: # # SELECT * FROM employees WHERE hire_date > '1987-09-07' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Which employees were hired after September 7th, 1987? # ### SQL: # # SELECT * FROM employees WHERE hire_date > '1987-09-07' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the job title of jobs which minimum salary is greater than 9000. # ### SQL: # # SELECT job_title FROM jobs WHERE min_salary > 9000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Which job titles correspond to jobs with salaries over 9000? # ### SQL: # # SELECT job_title FROM jobs WHERE min_salary > 9000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display job Title, the difference between minimum and maximum salaries for those jobs which max salary within the range 12000 to 18000. # ### SQL: # # SELECT job_title , max_salary - min_salary FROM jobs WHERE max_salary BETWEEN 12000 AND 18000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the job titles, and range of salaries for jobs with maximum salary between 12000 and 18000? # ### SQL: # # SELECT job_title , max_salary - min_salary FROM jobs WHERE max_salary BETWEEN 12000 AND 18000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the emails of the employees who have no commission percentage and salary within the range 7000 to 12000 and works in that department which number is 50. # ### SQL: # # SELECT email FROM employees WHERE commission_pct = "null" AND salary BETWEEN 7000 AND 12000 AND department_id = 50 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the emails of employees with null commission, salary between 7000 and 12000, and who work in department 50? # ### SQL: # # SELECT email FROM employees WHERE commission_pct = "null" AND salary BETWEEN 7000 AND 12000 AND department_id = 50 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the employee ID for each employee and the date on which he ended his previous job. # ### SQL: # # SELECT employee_id , MAX(end_date) FROM job_history GROUP BY employee_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the employee ids for each employee and final dates of employment at their last job? # ### SQL: # # SELECT employee_id , MAX(end_date) FROM job_history GROUP BY employee_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display those departments where more than ten employees work who got a commission percentage. # ### SQL: # # SELECT department_id FROM employees GROUP BY department_id HAVING COUNT(commission_pct) > 10 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the department ids for which more than 10 employees had a commission? # ### SQL: # # SELECT department_id FROM employees GROUP BY department_id HAVING COUNT(commission_pct) > 10 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the ids of the departments where any manager is managing 4 or more employees. # ### SQL: # # SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are department ids for departments with managers managing more than 3 employees? # ### SQL: # # SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the average salary of employees for each department who gets a commission percentage. # ### SQL: # # SELECT department_id , AVG(salary) FROM employees WHERE commission_pct != "null" GROUP BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is the average salary of employees who have a commission percentage that is not null? # ### SQL: # # SELECT department_id , AVG(salary) FROM employees WHERE commission_pct != "null" GROUP BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the country ID and number of cities for each country. # ### SQL: # # SELECT country_id , COUNT(*) FROM locations GROUP BY country_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Give the country id and corresponding count of cities in each country. # ### SQL: # # SELECT country_id , COUNT(*) FROM locations GROUP BY country_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display job ID for those jobs that were done by two or more for more than 300 days. # ### SQL: # # SELECT job_id FROM job_history WHERE end_date - start_date > 300 GROUP BY job_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the job ids for jobs done more than once for a period of more than 300 days? # ### SQL: # # SELECT job_id FROM job_history WHERE end_date - start_date > 300 GROUP BY job_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the ID for those employees who did two or more jobs in the past. # ### SQL: # # SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the employee ids for employees who have held two or more jobs? # ### SQL: # # SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find employee with ID and name of the country presently where (s)he is working. # ### SQL: # # SELECT T1.employee_id , T4.country_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id JOIN locations AS T3 ON T2.location_id = T3.location_id JOIN countries AS T4 ON T3.country_id = T4.country_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are all the employee ids and the names of the countries in which they work? # ### SQL: # # SELECT T1.employee_id , T4.country_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id JOIN locations AS T3 ON T2.location_id = T3.location_id JOIN countries AS T4 ON T3.country_id = T4.country_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the department name and number of employees in each of the department. # ### SQL: # # SELECT T2.department_name , COUNT(*) FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id GROUP BY T2.department_name # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Give the name of each department and the number of employees in each. # ### SQL: # # SELECT T2.department_name , COUNT(*) FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id GROUP BY T2.department_name # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Can you return all detailed info of jobs which was done by any of the employees who is presently earning a salary on and above 12000? # ### SQL: # # SELECT * FROM job_history AS T1 JOIN employees AS T2 ON T1.employee_id = T2.employee_id WHERE T2.salary >= 12000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the job history info done by employees earning a salary greater than or equal to 12000? # ### SQL: # # SELECT * FROM job_history AS T1 JOIN employees AS T2 ON T1.employee_id = T2.employee_id WHERE T2.salary >= 12000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display job title and average salary of employees. # ### SQL: # # SELECT job_title , AVG(salary) FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id GROUP BY T2.job_title # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is the average salary for each job title? # ### SQL: # # SELECT job_title , AVG(salary) FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id GROUP BY T2.job_title # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is the full name ( first name and last name ) for those employees who gets more salary than the employee whose id is 163? # ### SQL: # # SELECT first_name , last_name FROM employees WHERE salary > (SELECT salary FROM employees WHERE employee_id = 163 ) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Provide the full names of employees earning more than the employee with id 163. # ### SQL: # # SELECT first_name , last_name FROM employees WHERE salary > (SELECT salary FROM employees WHERE employee_id = 163 ) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # return the smallest salary for every departments. # ### SQL: # # SELECT MIN(salary) , department_id FROM employees GROUP BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is the minimum salary in each department? # ### SQL: # # SELECT MIN(salary) , department_id FROM employees GROUP BY department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the first name and last name and department id for those employees who earn such amount of salary which is the smallest salary of any of the departments. # ### SQL: # # SELECT first_name , last_name , department_id FROM employees WHERE salary IN (SELECT MIN(salary) FROM employees GROUP BY department_id) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the full names and department ids for the lowest paid employees across all departments. # ### SQL: # # SELECT first_name , last_name , department_id FROM employees WHERE salary IN (SELECT MIN(salary) FROM employees GROUP BY department_id) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the employee id for all employees who earn more than the average salary. # ### SQL: # # SELECT employee_id FROM employees WHERE salary > (SELECT AVG(salary) FROM employees) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the employee ids for employees who make more than the average? # ### SQL: # # SELECT employee_id FROM employees WHERE salary > (SELECT AVG(salary) FROM employees) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the employee id and salary of all employees who report to Payam (first name). # ### SQL: # # SELECT employee_id , salary FROM employees WHERE manager_id = (SELECT employee_id FROM employees WHERE first_name = 'Payam' ) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the employee ids of employees who report to Payam, and what are their salaries? # ### SQL: # # SELECT employee_id , salary FROM employees WHERE manager_id = (SELECT employee_id FROM employees WHERE first_name = 'Payam' ) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # find the name of all departments that do actually have one or more employees assigned to them. # ### SQL: # # SELECT DISTINCT T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the names of departments that have at least one employee. # ### SQL: # # SELECT DISTINCT T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # get the details of employees who manage a department. # ### SQL: # # SELECT DISTINCT * FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T1.employee_id = T2.manager_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information regarding employees who are managers? # ### SQL: # # SELECT DISTINCT * FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T1.employee_id = T2.manager_id # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display all the information about the department Marketing. # ### SQL: # # SELECT * FROM departments WHERE department_name = 'Marketing' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information about the Marketing department? # ### SQL: # # SELECT * FROM departments WHERE department_name = 'Marketing' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the ID for those employees who did two or more jobs in the past. # ### SQL: # # SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the employee ids for those who had two or more jobs. # ### SQL: # # SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the unique ids of those departments where any manager is managing 4 or more employees. # ### SQL: # # SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Give the distinct department ids of departments in which a manager is in charge of 4 or more employees? # ### SQL: # # SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the job ID for those jobs which average salary is above 8000. # ### SQL: # # SELECT job_id FROM employees GROUP BY job_id HAVING AVG(salary) > 8000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the job ids corresponding to jobs with average salary above 8000? # ### SQL: # # SELECT job_id FROM employees GROUP BY job_id HAVING AVG(salary) > 8000 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the employee ID and job name for all those jobs in department 80. # ### SQL: # # SELECT T1.employee_id , T2.job_title FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id WHERE T1.department_id = 80 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # what are the employee ids and job titles for employees in department 80? # ### SQL: # # SELECT T1.employee_id , T2.job_title FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id WHERE T1.department_id = 80 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is the first name and job id for all employees in the Finance department? # ### SQL: # # SELECT T1.first_name , T1.job_id FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T2.department_name = 'Finance' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Give the first name and job id for all employees in the Finance department. # ### SQL: # # SELECT T1.first_name , T1.job_id FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T2.department_name = 'Finance' # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display all the information of the employees whose salary if within the range of smallest salary and 2500. # ### SQL: # # SELECT * FROM employees WHERE salary BETWEEN (SELECT MIN(salary) FROM employees) AND 2500 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What is all the information regarding employees with salaries above the minimum and under 2500? # ### SQL: # # SELECT * FROM employees WHERE salary BETWEEN (SELECT MIN(salary) FROM employees) AND 2500 # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # Find the ids of the employees who does not work in those departments where some employees works whose manager id within the range 100 and 200. # ### SQL: # # SELECT * FROM employees WHERE department_id NOT IN (SELECT department_id FROM departments WHERE manager_id BETWEEN 100 AND 200) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # What are the ids for employees who do not work in departments with managers that have ids between 100 and 200? # ### SQL: # # SELECT * FROM employees WHERE department_id NOT IN (SELECT department_id FROM departments WHERE manager_id BETWEEN 100 AND 200) # ### End.
hr_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # regions ( REGION_ID, REGION_NAME ) # countries ( COUNTRY_ID, COUNTRY_NAME, REGION_ID ) # departments ( DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID ) # jobs ( JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY ) # employees ( EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID ) # job_history ( EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID ) # locations ( LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID ) # # countries.REGION_ID can be joined with regions.REGION_ID # employees.JOB_ID can be joined with jobs.JOB_ID # employees.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.JOB_ID can be joined with jobs.JOB_ID # job_history.DEPARTMENT_ID can be joined with departments.DEPARTMENT_ID # job_history.EMPLOYEE_ID can be joined with employees.EMPLOYEE_ID # locations.COUNTRY_ID can be joined with countries.COUNTRY_ID # ### Question: # # display the employee name ( first name and last name ) and hire date for all employees in the same department as Clara. # ### SQL: # # SELECT first_name , last_name , hire_date FROM employees WHERE department_id = (SELECT department_id FROM employees WHERE first_name = "Clara") # ### End.

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