Split (2)

train · 8.66k rows

db_name stringclasses 146 values	prompt stringlengths 310 4.81k
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the name of companies whose revenue is greater than the average revenue of all companies. # ### SQL: # # SELECT name FROM manufacturers WHERE revenue > (SELECT avg(revenue) FROM manufacturers) # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names of manufacturers with revenue greater than the average of all revenues? # ### SQL: # # SELECT name FROM manufacturers WHERE revenue > (SELECT avg(revenue) FROM manufacturers) # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the name of companies whose revenue is smaller than the revenue of all companies based in Austin. # ### SQL: # # SELECT name FROM manufacturers WHERE revenue < (SELECT min(revenue) FROM manufacturers WHERE headquarter = 'Austin') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names of companies with revenue less than the lowest revenue of any manufacturer in Austin? # ### SQL: # # SELECT name FROM manufacturers WHERE revenue < (SELECT min(revenue) FROM manufacturers WHERE headquarter = 'Austin') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the total revenue of companies whose revenue is larger than the revenue of some companies based in Austin. # ### SQL: # # SELECT sum(revenue) FROM manufacturers WHERE revenue > (SELECT min(revenue) FROM manufacturers WHERE headquarter = 'Austin') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the total revenue of companies with revenue greater than the lowest revenue of any manufacturer in Austin? # ### SQL: # # SELECT sum(revenue) FROM manufacturers WHERE revenue > (SELECT min(revenue) FROM manufacturers WHERE headquarter = 'Austin') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the total revenue of companies of each founder. # ### SQL: # # SELECT sum(revenue) , founder FROM manufacturers GROUP BY founder # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the total revenue of companies started by founder? # ### SQL: # # SELECT sum(revenue) , founder FROM manufacturers GROUP BY founder # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the name and revenue of the company that earns the highest revenue in each city. # ### SQL: # # SELECT name , max(revenue) , Headquarter FROM manufacturers GROUP BY Headquarter # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names and revenues of the companies with the highest revenues in each headquarter city? # ### SQL: # # SELECT name , max(revenue) , Headquarter FROM manufacturers GROUP BY Headquarter # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the total revenue for each manufacturer. # ### SQL: # # SELECT sum(revenue) , name FROM manufacturers GROUP BY name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the total revenue of each manufacturer? # ### SQL: # # SELECT sum(revenue) , name FROM manufacturers GROUP BY name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the average prices of all products from each manufacture, and list each company's name. # ### SQL: # # SELECT avg(T1.price) , T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the average prices of products for each manufacturer? # ### SQL: # # SELECT avg(T1.price) , T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the number of different products that are produced by companies at different headquarter cities. # ### SQL: # # SELECT count(DISTINCT T1.name) , T2.Headquarter FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.Headquarter # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # How many different products are produced in each headquarter city? # ### SQL: # # SELECT count(DISTINCT T1.name) , T2.Headquarter FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.Headquarter # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find number of products which Sony does not make. # ### SQL: # # SELECT count(DISTINCT name) FROM products WHERE name NOT IN (SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # How many products are not made by Sony? # ### SQL: # # SELECT count(DISTINCT name) FROM products WHERE name NOT IN (SELECT T1.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T2.name = 'Sony') # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the name of companies that do not make DVD drive. # ### SQL: # # SELECT name FROM manufacturers EXCEPT SELECT T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T1.name = 'DVD drive' # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names of companies that do not make DVD drives? # ### SQL: # # SELECT name FROM manufacturers EXCEPT SELECT T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T1.name = 'DVD drive' # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find the number of products for each manufacturer, showing the name of each company. # ### SQL: # # SELECT count(*) , T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # How many products are there for each manufacturer? # ### SQL: # # SELECT count(*) , T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the names of all the products in the store. # ### SQL: # # SELECT Name FROM Products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names of all products? # ### SQL: # # SELECT Name FROM Products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the names and the prices of all the products in the store. # ### SQL: # # SELECT name , price FROM products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names and prices of all products in the store? # ### SQL: # # SELECT name , price FROM products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the name of the products with a price less than or equal to $200. # ### SQL: # # SELECT name FROM products WHERE price <= 200 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names of products with price at most 200? # ### SQL: # # SELECT name FROM products WHERE price <= 200 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Find all information of all the products with a price between $60 and $120. # ### SQL: # # SELECT * FROM products WHERE price BETWEEN 60 AND 120 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is all the information of all the products that have a price between 60 and 120? # ### SQL: # # SELECT * FROM products WHERE price BETWEEN 60 AND 120 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Compute the average price of all the products. # ### SQL: # # SELECT avg(price) FROM products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the average price across all products? # ### SQL: # # SELECT avg(price) FROM products # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Compute the average price of all products with manufacturer code equal to 2. # ### SQL: # # SELECT avg(price) FROM products WHERE Manufacturer = 2 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the average price of products with manufacturer codes equal to 2? # ### SQL: # # SELECT avg(price) FROM products WHERE Manufacturer = 2 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Compute the number of products with a price larger than or equal to $180. # ### SQL: # # SELECT count(*) FROM products WHERE price >= 180 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # How many products have prices of at least 180? # ### SQL: # # SELECT count(*) FROM products WHERE price >= 180 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the name and price of all products with a price larger than or equal to $180, and sort first by price (in descending order), and then by name (in ascending order). # ### SQL: # # SELECT name , price FROM products WHERE price >= 180 ORDER BY price DESC , name ASC # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names and prices of products that cost at least 180, sorted by price decreasing and name ascending? # ### SQL: # # SELECT name , price FROM products WHERE price >= 180 ORDER BY price DESC , name ASC # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select all the data from the products and each product's manufacturer. # ### SQL: # # SELECT * FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is all the product data, as well as each product's manufacturer? # ### SQL: # # SELECT * FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the average price of each manufacturer's products, showing only the manufacturer's code. # ### SQL: # # SELECT AVG(Price) , Manufacturer FROM Products GROUP BY Manufacturer # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the average prices of products, grouped by manufacturer code? # ### SQL: # # SELECT AVG(Price) , Manufacturer FROM Products GROUP BY Manufacturer # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the average price of each manufacturer's products, showing the manufacturer's name. # ### SQL: # # SELECT avg(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the average prices of products, grouped by manufacturer name? # ### SQL: # # SELECT avg(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the names of manufacturer whose products have an average price higher than or equal to $150. # ### SQL: # # SELECT avg(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name HAVING avg(T1.price) >= 150 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the names and average prices of products for manufacturers whose products cost on average 150 or more? # ### SQL: # # SELECT avg(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name HAVING avg(T1.price) >= 150 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the name and price of the cheapest product. # ### SQL: # # SELECT name , price FROM Products ORDER BY price ASC LIMIT 1 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What is the name and price of the cheapest product? # ### SQL: # # SELECT name , price FROM Products ORDER BY price ASC LIMIT 1 # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the name of each manufacturer along with the name and price of its most expensive product. # ### SQL: # # SELECT T1.Name , max(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # For each manufacturer name, what are the names and prices of their most expensive product? # ### SQL: # # SELECT T1.Name , max(T1.Price) , T2.name FROM products AS T1 JOIN Manufacturers AS T2 ON T1.manufacturer = T2.code GROUP BY T2.name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # Select the code of the product that is cheapest in each product category. # ### SQL: # # SELECT code , name , min(price) FROM products GROUP BY name # ### End.
manufactory_1	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Manufacturers ( Code, Name, Headquarter, Founder, Revenue ) # Products ( Code, Name, Price, Manufacturer ) # # Products.Manufacturer can be joined with Manufacturers.Code # ### Question: # # What are the codes and names of the cheapest products in each category? # ### SQL: # # SELECT code , name , min(price) FROM products GROUP BY name # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What is the id of the problem log that is created most recently? # ### SQL: # # SELECT problem_log_id FROM problem_log ORDER BY log_entry_date DESC LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problem log was created most recently? Give me the log id. # ### SQL: # # SELECT problem_log_id FROM problem_log ORDER BY log_entry_date DESC LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What is the oldest log id and its corresponding problem id? # ### SQL: # # SELECT problem_log_id , problem_id FROM problem_log ORDER BY log_entry_date LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the oldest log id and its corresponding problem id. # ### SQL: # # SELECT problem_log_id , problem_id FROM problem_log ORDER BY log_entry_date LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find all the ids and dates of the logs for the problem whose id is 10. # ### SQL: # # SELECT problem_log_id , log_entry_date FROM problem_log WHERE problem_id = 10 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # For the problem with id 10, return the ids and dates of its problem logs. # ### SQL: # # SELECT problem_log_id , log_entry_date FROM problem_log WHERE problem_id = 10 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List all the log ids and their descriptions from the problem logs. # ### SQL: # # SELECT problem_log_id , log_entry_description FROM problem_log # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the log id and entry description of each problem? # ### SQL: # # SELECT problem_log_id , log_entry_description FROM problem_log # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List the first and last names of all distinct staff members who are assigned to the problem whose id is 1. # ### SQL: # # SELECT DISTINCT staff_first_name , staff_last_name FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T2.problem_id = 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which staff members are assigned to the problem with id 1? Give me their first and last names. # ### SQL: # # SELECT DISTINCT staff_first_name , staff_last_name FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T2.problem_id = 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List the problem id and log id which are assigned to the staff named Rylan Homenick. # ### SQL: # # SELECT DISTINCT T2.problem_id , T2.problem_log_id FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T1.staff_first_name = "Rylan" AND T1.staff_last_name = "Homenick" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problem id and log id are assigned to the staff named Rylan Homenick? # ### SQL: # # SELECT DISTINCT T2.problem_id , T2.problem_log_id FROM staff AS T1 JOIN problem_log AS T2 ON T1.staff_id = T2.assigned_to_staff_id WHERE T1.staff_first_name = "Rylan" AND T1.staff_last_name = "Homenick" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # How many problems are there for product voluptatem? # ### SQL: # # SELECT count(*) FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id WHERE T1.product_name = "voluptatem" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # How many problems did the product called "voluptatem" have in record? # ### SQL: # # SELECT count(*) FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id WHERE T1.product_name = "voluptatem" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # How many problems does the product with the most problems have? List the number of the problems and product name. # ### SQL: # # SELECT count() , T1.product_name FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id GROUP BY T1.product_name ORDER BY count() DESC LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which product has the most problems? Give me the number of problems and the product name. # ### SQL: # # SELECT count() , T1.product_name FROM product AS T1 JOIN problems AS T2 ON T1.product_id = T2.product_id GROUP BY T1.product_name ORDER BY count() DESC LIMIT 1 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Give me a list of descriptions of the problems that are reported by the staff whose first name is Christop. # ### SQL: # # SELECT T1.problem_description FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problems are reported by the staff with first name "Christop"? Show the descriptions of the problems. # ### SQL: # # SELECT T1.problem_description FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the ids of the problems that are reported by the staff whose last name is Bosco. # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_last_name = "Bosco" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problems are reported by the staff with last name "Bosco"? Show the ids of the problems. # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_last_name = "Bosco" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the ids of the problems which are reported after 1978-06-26? # ### SQL: # # SELECT problem_id FROM problems WHERE date_problem_reported > "1978-06-26" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the ids of the problems reported after 1978-06-26. # ### SQL: # # SELECT problem_id FROM problems WHERE date_problem_reported > "1978-06-26" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the ids of the problems which are reported before 1978-06-26? # ### SQL: # # SELECT problem_id FROM problems WHERE date_problem_reported < "1978-06-26" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problems are reported before 1978-06-26? Give me the ids of the problems. # ### SQL: # # SELECT problem_id FROM problems WHERE date_problem_reported < "1978-06-26" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # For each product which has problems, what are the number of problems and the product id? # ### SQL: # # SELECT count(*) , T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # For each product with some problems, list the count of problems and the product id. # ### SQL: # # SELECT count(*) , T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # For each product that has problems, find the number of problems reported after 1986-11-13 and the product id? # ### SQL: # # SELECT count(*) , T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T1.date_problem_reported > "1986-11-13" GROUP BY T2.product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the products that have problems reported after 1986-11-13? Give me the product id and the count of problems reported after 1986-11-13. # ### SQL: # # SELECT count(*) , T2.product_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T1.date_problem_reported > "1986-11-13" GROUP BY T2.product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List the names of all the distinct product names in alphabetical order? # ### SQL: # # SELECT DISTINCT product_name FROM product ORDER BY product_name # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Sort all the distinct product names in alphabetical order. # ### SQL: # # SELECT DISTINCT product_name FROM product ORDER BY product_name # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List all the distinct product names ordered by product id? # ### SQL: # # SELECT DISTINCT product_name FROM product ORDER BY product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What is the list of distinct product names sorted by product id? # ### SQL: # # SELECT DISTINCT product_name FROM product ORDER BY product_id # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the id of problems reported by the staff named Dameon Frami or Jolie Weber? # ### SQL: # # SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Dameon" AND T2.staff_last_name = "Frami" UNION SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Jolie" AND T2.staff_last_name = "Weber" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problems were reported by the staff named Dameon Frami or Jolie Weber? Give me the ids of the problems. # ### SQL: # # SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Dameon" AND T2.staff_last_name = "Frami" UNION SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Jolie" AND T2.staff_last_name = "Weber" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the product ids for the problems reported by Christop Berge with closure authorised by Ashley Medhurst? # ### SQL: # # SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" AND T2.staff_last_name = "Berge" INTERSECT SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.closure_authorised_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Ashley" AND T2.staff_last_name = "Medhurst" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # For which product was there a problem reported by Christop Berge, with closure authorised by Ashley Medhurst? Return the product ids. # ### SQL: # # SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Christop" AND T2.staff_last_name = "Berge" INTERSECT SELECT product_id FROM problems AS T1 JOIN staff AS T2 ON T1.closure_authorised_by_staff_id = T2.staff_id WHERE T2.staff_first_name = "Ashley" AND T2.staff_last_name = "Medhurst" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the ids of the problems reported before the date of any problem reported by Lysanne Turcotte? # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported < ( SELECT min(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Lysanne" AND T4.staff_last_name = "Turcotte" ) # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which problems were reported before the date of any problem reported by the staff Lysanne Turcotte? Give me the ids of the problems. # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported < ( SELECT min(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Lysanne" AND T4.staff_last_name = "Turcotte" ) # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the ids of the problems reported after the date of any problems reported by Rylan Homenick? # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported > ( SELECT max(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Rylan" AND T4.staff_last_name = "Homenick" ) # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the ids of the problems reported after the date of any problems reported by the staff Rylan Homenick. # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN staff AS T2 ON T1.reported_by_staff_id = T2.staff_id WHERE date_problem_reported > ( SELECT max(date_problem_reported) FROM problems AS T3 JOIN staff AS T4 ON T3.reported_by_staff_id = T4.staff_id WHERE T4.staff_first_name = "Rylan" AND T4.staff_last_name = "Homenick" ) # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the top 3 products which have the largest number of problems? # ### SQL: # # SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_name ORDER BY count(*) DESC LIMIT 3 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the three products that have the most problems?s # ### SQL: # # SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_name ORDER BY count(*) DESC LIMIT 3 # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # List the ids of the problems from the product "voluptatem" that are reported after 1995? # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T2.product_name = "voluptatem" AND T1.date_problem_reported > "1995" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # What are the ids of the problems that are from the product "voluptatem" and are reported after 1995? # ### SQL: # # SELECT T1.problem_id FROM problems AS T1 JOIN product AS T2 ON T1.product_id = T2.product_id WHERE T2.product_name = "voluptatem" AND T1.date_problem_reported > "1995" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the first and last name of the staff members who reported problems from the product "rem" but not "aut"? # ### SQL: # # SELECT T3.staff_first_name , T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "rem" EXCEPT SELECT T3.staff_first_name , T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "aut" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which staff members who reported problems from the product "rem" but not "aut"? Give me their first and last names. # ### SQL: # # SELECT T3.staff_first_name , T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "rem" EXCEPT SELECT T3.staff_first_name , T3.staff_last_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T2.product_name = "aut" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Find the products which have problems reported by both Lacey Bosco and Kenton Champlin? # ### SQL: # # SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Lacey" AND T3.staff_last_name = "Bosco" INTERSECT SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Kenton" AND T3.staff_last_name = "Champlin" # ### End.
tracking_software_problems	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Problem_Category_Codes ( problem_category_code, problem_category_description ) # Problem_Log ( problem_log_id, assigned_to_staff_id, problem_id, problem_category_code, problem_status_code, log_entry_date, log_entry_description, log_entry_fix, other_log_details ) # Problem_Status_Codes ( problem_status_code, problem_status_description ) # Product ( product_id, product_name, product_details ) # Staff ( staff_id, staff_first_name, staff_last_name, other_staff_details ) # Problems ( problem_id, product_id, closure_authorised_by_staff_id, reported_by_staff_id, date_problem_reported, date_problem_closed, problem_description, other_problem_details ) # # Problem_Log.problem_status_code can be joined with Problem_Status_Codes.problem_status_code # Problem_Log.problem_id can be joined with Problems.problem_id # Problem_Log.assigned_to_staff_id can be joined with Staff.staff_id # Problem_Log.problem_category_code can be joined with Problem_Category_Codes.problem_category_code # Problems.reported_by_staff_id can be joined with Staff.staff_id # Problems.product_id can be joined with Product.product_id # Problems.closure_authorised_by_staff_id can be joined with Staff.staff_id # ### Question: # # Which products have problems reported by both the staff named Lacey Bosco and the staff named Kenton Champlin? # ### SQL: # # SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Lacey" AND T3.staff_last_name = "Bosco" INTERSECT SELECT T2.product_name FROM problems AS T1 JOIN product AS T2 JOIN staff AS T3 ON T1.product_id = T2.product_id AND T1.reported_by_staff_id = T3.staff_id WHERE T3.staff_first_name = "Kenton" AND T3.staff_last_name = "Champlin" # ### End.

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