Split (2)

train · 8.66k rows

db_name stringclasses 146 values	prompt stringlengths 310 4.81k
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the name, location, and number of platforms for all stations. # ### SQL: # # SELECT name , LOCATION , number_of_platforms FROM station # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # What are all locations of train stations? # ### SQL: # # SELECT DISTINCT LOCATION FROM station # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the names and total passengers for all train stations not in London. # ### SQL: # # SELECT name , total_passengers FROM station WHERE LOCATION != 'London' # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the names and main services for train stations that have the top three total number of passengers. # ### SQL: # # SELECT name , main_services FROM station ORDER BY total_passengers DESC LIMIT 3 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # What is the average and maximum number of total passengers for train stations in London or Glasgow? # ### SQL: # # SELECT avg(total_passengers) , max(total_passengers) FROM station WHERE LOCATION = 'London' OR LOCATION = 'Glasgow' # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show all locations and the total number of platforms and passengers for all train stations in each location. # ### SQL: # # SELECT LOCATION , sum(number_of_platforms) , sum(total_passengers) FROM station GROUP BY LOCATION # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show all locations that have train stations with at least 15 platforms and train stations with more than 25 total passengers. # ### SQL: # # SELECT DISTINCT LOCATION FROM station WHERE number_of_platforms >= 15 AND total_passengers > 25 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show all locations which don't have a train station with at least 15 platforms. # ### SQL: # # SELECT LOCATION FROM station EXCEPT SELECT LOCATION FROM station WHERE number_of_platforms >= 15 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the location with most number of train stations. # ### SQL: # # SELECT LOCATION FROM station GROUP BY LOCATION ORDER BY count(*) DESC LIMIT 1 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the name, time, and service for all trains. # ### SQL: # # SELECT name , TIME , service FROM train # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the number of trains # ### SQL: # # SELECT count(*) FROM train # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the name and service for all trains in order by time. # ### SQL: # # SELECT name , service FROM train ORDER BY TIME # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the station name and number of trains in each station. # ### SQL: # # SELECT T2.name , count(*) FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id GROUP BY T1.station_id # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # show the train name and station name for each train. # ### SQL: # # SELECT T2.name , T3.name FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id JOIN train AS T3 ON T3.train_id = T1.train_id # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show all train names and times in stations in London in descending order by train time. # ### SQL: # # SELECT T3.name , T3.time FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id JOIN train AS T3 ON T3.train_id = T1.train_id WHERE T2.location = 'London' ORDER BY T3.time DESC # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the station name with greatest number of trains. # ### SQL: # # SELECT T2.name FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id GROUP BY T1.station_id ORDER BY count(*) DESC LIMIT 1 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show the station name with at least two trains. # ### SQL: # # SELECT T2.name FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id GROUP BY T1.station_id HAVING count(*) >= 2 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show all locations with only 1 station. # ### SQL: # # SELECT LOCATION FROM station GROUP BY LOCATION HAVING count(*) = 1 # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Show station names without any trains. # ### SQL: # # SELECT name FROM station WHERE station_id NOT IN (SELECT station_id FROM train_station) # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # What are the names of the stations which serve both "Ananthapuri Express" and "Guruvayur Express" trains? # ### SQL: # # SELECT T2.name FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id JOIN train AS T3 ON T3.train_id = T1.train_id WHERE T3.Name = "Ananthapuri Express" INTERSECT SELECT T2.name FROM train_station AS T1 JOIN station AS T2 ON T1.station_id = T2.station_id JOIN train AS T3 ON T3.train_id = T1.train_id WHERE T3.Name = "Guruvayur Express" # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # Find the names of the trains that do not pass any station located in London. # ### SQL: # # SELECT T2.name FROM train_station AS T1 JOIN train AS T2 ON T1.train_id = T2.train_id WHERE T1.station_id NOT IN (SELECT T4.station_id FROM train_station AS T3 JOIN station AS T4 ON T3.station_id = T4.station_id WHERE t4.location = "London") # ### End.
train_station	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # station ( Station_ID, Name, Annual_entry_exit, Annual_interchanges, Total_Passengers, Location, Main_Services, Number_of_Platforms ) # train ( Train_ID, Name, Time, Service ) # train_station ( Train_ID, Station_ID ) # # train_station.Station_ID can be joined with station.Station_ID # train_station.Train_ID can be joined with train.Train_ID # ### Question: # # List the names and locations of all stations ordered by their yearly entry exit and interchange amounts. # ### SQL: # # SELECT name , LOCATION FROM station ORDER BY Annual_entry_exit , Annual_interchanges # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List all vehicle id # ### SQL: # # SELECT vehicle_id FROM Vehicles; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the ids of all vehicles? # ### SQL: # # SELECT vehicle_id FROM Vehicles; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many vehicle in total? # ### SQL: # # SELECT count(*) FROM Vehicles; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many vehicles exist? # ### SQL: # # SELECT count(*) FROM Vehicles; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Show the detail of vehicle with id 1. # ### SQL: # # SELECT vehicle_details FROM Vehicles WHERE vehicle_id = 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the details of the car with id 1? # ### SQL: # # SELECT vehicle_details FROM Vehicles WHERE vehicle_id = 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List the first name middle name and last name of all staff. # ### SQL: # # SELECT first_name , middle_name , last_name FROM Staff; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the first, middle, and last names of all staff? # ### SQL: # # SELECT first_name , middle_name , last_name FROM Staff; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the birthday of the staff member with first name as Janessa and last name as Sawayn? # ### SQL: # # SELECT date_of_birth FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the date of birth for the staff member named Janessa Sawayn? # ### SQL: # # SELECT date_of_birth FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did the staff member with first name as Janessa and last name as Sawayn join the company? # ### SQL: # # SELECT date_joined_staff FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did the staff member named Janessa Sawayn join the company? # ### SQL: # # SELECT date_joined_staff FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did the staff member with first name as Janessa and last name as Sawayn leave the company? # ### SQL: # # SELECT date_left_staff FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did the staff member Janessa Sawayn leave the company? # ### SQL: # # SELECT date_left_staff FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many staff have the first name Ludie? # ### SQL: # # SELECT count(*) FROM Staff WHERE first_name = "Ludie"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many employees have a first name of Ludie? # ### SQL: # # SELECT count(*) FROM Staff WHERE first_name = "Ludie"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the nickname of staff with first name as Janessa and last name as Sawayn? # ### SQL: # # SELECT nickname FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the nickname of the employee named Janessa Sawayn? # ### SQL: # # SELECT nickname FROM Staff WHERE first_name = "Janessa" AND last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many staff in total? # ### SQL: # # SELECT count(*) FROM Staff; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many employees are there? # ### SQL: # # SELECT count(*) FROM Staff; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which city does staff with first name as Janessa and last name as Sawayn live? # ### SQL: # # SELECT T1.city FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # In what city does Janessa Sawayn live? # ### SQL: # # SELECT T1.city FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which country and state does staff with first name as Janessa and last name as Sawayn lived? # ### SQL: # # SELECT T1.country , T1.state_province_county FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # In which country and state does Janessa Sawayn live? # ### SQL: # # SELECT T1.country , T1.state_province_county FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How long is the total lesson time took by customer with first name as Rylan and last name as Goodwin? # ### SQL: # # SELECT sum(T1.lesson_time) FROM Lessons AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Rylan" AND T2.last_name = "Goodwin"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How long is the total lesson time took by the customer named Rylan Goodwin? # ### SQL: # # SELECT sum(T1.lesson_time) FROM Lessons AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Rylan" AND T2.last_name = "Goodwin"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the zip code of staff with first name as Janessa and last name as Sawayn lived? # ### SQL: # # SELECT T1.zip_postcode FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the zip code of the hosue of the employee named Janessa Sawayn? # ### SQL: # # SELECT T1.zip_postcode FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T2.first_name = "Janessa" AND T2.last_name = "Sawayn"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many staff live in state Georgia? # ### SQL: # # SELECT count(*) FROM Addresses WHERE state_province_county = "Georgia"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many employees live in Georgia? # ### SQL: # # SELECT count(*) FROM Addresses WHERE state_province_county = "Georgia"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Find out the first name and last name of staff lived in city Damianfort. # ### SQL: # # SELECT T2.first_name , T2.last_name FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T1.city = "Damianfort"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the first and last name of all employees who live in the city Damianfort? # ### SQL: # # SELECT T2.first_name , T2.last_name FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id WHERE T1.city = "Damianfort"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which city lives most of staffs? List the city name and number of staffs. # ### SQL: # # SELECT T1.city , count() FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id GROUP BY T1.city ORDER BY count() DESC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # In which city do the most employees live and how many of them live there? # ### SQL: # # SELECT T1.city , count() FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id GROUP BY T1.city ORDER BY count() DESC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List the states which have between 2 to 4 staffs living there. # ### SQL: # # SELECT T1.state_province_county FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id GROUP BY T1.state_province_county HAVING count(*) BETWEEN 2 AND 4; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the names of the states that have 2 to 4 employees living there? # ### SQL: # # SELECT T1.state_province_county FROM Addresses AS T1 JOIN Staff AS T2 ON T1.address_id = T2.staff_address_id GROUP BY T1.state_province_county HAVING count(*) BETWEEN 2 AND 4; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List the first name and last name of all customers. # ### SQL: # # SELECT first_name , last_name FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the first and last names for all customers? # ### SQL: # # SELECT first_name , last_name FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List email address and birthday of customer whose first name as Carole. # ### SQL: # # SELECT email_address , date_of_birth FROM Customers WHERE first_name = "Carole" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the email addresses and date of births for all customers who have a first name of Carole? # ### SQL: # # SELECT email_address , date_of_birth FROM Customers WHERE first_name = "Carole" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List phone number and email address of customer with more than 2000 outstanding balance. # ### SQL: # # SELECT phone_number , email_address FROM Customers WHERE amount_outstanding > 2000; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the phone numbers and email addresses of all customers who have an outstanding balance of more than 2000? # ### SQL: # # SELECT phone_number , email_address FROM Customers WHERE amount_outstanding > 2000; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the status code, mobile phone number and email address of customer with last name as Kohler or first name as Marina? # ### SQL: # # SELECT customer_status_code , cell_mobile_phone_number , email_address FROM Customers WHERE first_name = "Marina" OR last_name = "Kohler" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the status code, phone number, and email address of the customer whose last name is Kohler or whose first name is Marina? # ### SQL: # # SELECT customer_status_code , cell_mobile_phone_number , email_address FROM Customers WHERE first_name = "Marina" OR last_name = "Kohler" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When are the birthdays of customer who are classified as 'Good Customer' status? # ### SQL: # # SELECT date_of_birth FROM Customers WHERE customer_status_code = 'Good Customer' # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the date of birth of every customer whose status code is 'Good Customer'? # ### SQL: # # SELECT date_of_birth FROM Customers WHERE customer_status_code = 'Good Customer' # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did customer with first name as Carole and last name as Bernhard became a customer? # ### SQL: # # SELECT date_became_customer FROM Customers WHERE first_name = "Carole" AND last_name = "Bernhard"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # When did Carole Bernhard first become a customer? # ### SQL: # # SELECT date_became_customer FROM Customers WHERE first_name = "Carole" AND last_name = "Bernhard"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many customers in total? # ### SQL: # # SELECT count(*) FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many customers are there? # ### SQL: # # SELECT count(*) FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List all customer status codes and the number of customers having each status code. # ### SQL: # # SELECT customer_status_code , count(*) FROM Customers GROUP BY customer_status_code; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # For each customer status code, how many customers are classified that way? # ### SQL: # # SELECT customer_status_code , count(*) FROM Customers GROUP BY customer_status_code; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which customer status code has least number of customers? # ### SQL: # # SELECT customer_status_code FROM Customers GROUP BY customer_status_code ORDER BY count(*) ASC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the status code with the least number of customers? # ### SQL: # # SELECT customer_status_code FROM Customers GROUP BY customer_status_code ORDER BY count(*) ASC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many lessons taken by customer with first name as Rylan and last name as Goodwin were completed? # ### SQL: # # SELECT count(*) FROM Lessons AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Rylan" AND T2.last_name = "Goodwin" AND T1.lesson_status_code = "Completed"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many lessons did the customer Ryan Goodwin complete? # ### SQL: # # SELECT count(*) FROM Lessons AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Rylan" AND T2.last_name = "Goodwin" AND T1.lesson_status_code = "Completed"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is maximum, minimum and average amount of outstanding of customer? # ### SQL: # # SELECT max(amount_outstanding) , min(amount_outstanding) , avg(amount_outstanding) FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the maximum, minimum, and average amount of money outsanding for all customers? # ### SQL: # # SELECT max(amount_outstanding) , min(amount_outstanding) , avg(amount_outstanding) FROM Customers; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List the first name and last name of customers have the amount of outstanding between 1000 and 3000. # ### SQL: # # SELECT first_name , last_name FROM Customers WHERE amount_outstanding BETWEEN 1000 AND 3000; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the first and last names of all customers with between 1000 and 3000 dollars outstanding? # ### SQL: # # SELECT first_name , last_name FROM Customers WHERE amount_outstanding BETWEEN 1000 AND 3000; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List first name and last name of customers lived in city Lockmanfurt. # ### SQL: # # SELECT T1.first_name , T1.last_name FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T2.city = "Lockmanfurt"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the first and last names of all customers who lived in Lockmanfurt? # ### SQL: # # SELECT T1.first_name , T1.last_name FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T2.city = "Lockmanfurt"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which country does customer with first name as Carole and last name as Bernhard lived in? # ### SQL: # # SELECT T2.country FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T1.first_name = "Carole" AND T1.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the country in which the customer Carole Bernhard lived? # ### SQL: # # SELECT T2.country FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T1.first_name = "Carole" AND T1.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is zip code of customer with first name as Carole and last name as Bernhard? # ### SQL: # # SELECT T2.zip_postcode FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T1.first_name = "Carole" AND T1.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the zip code of the customer Carole Bernhard? # ### SQL: # # SELECT T2.zip_postcode FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id WHERE T1.first_name = "Carole" AND T1.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # Which city does has most number of customers? # ### SQL: # # SELECT T2.city FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id GROUP BY T2.city ORDER BY count(*) DESC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the city with the most customers? # ### SQL: # # SELECT T2.city FROM Customers AS T1 JOIN Addresses AS T2 ON T1.customer_address_id = T2.address_id GROUP BY T2.city ORDER BY count(*) DESC LIMIT 1; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How much in total does customer with first name as Carole and last name as Bernhard paid? # ### SQL: # # SELECT sum(T1.amount_payment) FROM Customer_Payments AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Carole" AND T2.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What is the total amount of moeny paid by the customer Carole Bernhard? # ### SQL: # # SELECT sum(T1.amount_payment) FROM Customer_Payments AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.first_name = "Carole" AND T2.last_name = "Bernhard" # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List the number of customers that did not have any payment history. # ### SQL: # # SELECT count(*) FROM Customers WHERE customer_id NOT IN ( SELECT customer_id FROM Customer_Payments ); # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many customers have no payment histories? # ### SQL: # # SELECT count(*) FROM Customers WHERE customer_id NOT IN ( SELECT customer_id FROM Customer_Payments ); # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List first name and last name of customers that have more than 2 payments. # ### SQL: # # SELECT T2.first_name , T2.last_name FROM Customer_Payments AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING count(*) > 2; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # What are the first and last names of all customers with more than 2 payments? # ### SQL: # # SELECT T2.first_name , T2.last_name FROM Customer_Payments AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING count(*) > 2; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # List all payment methods and number of payments using each payment methods. # ### SQL: # # SELECT payment_method_code , count(*) FROM Customer_Payments GROUP BY payment_method_code; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # For each payment method, how many payments were made? # ### SQL: # # SELECT payment_method_code , count(*) FROM Customer_Payments GROUP BY payment_method_code; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many lessons were in cancelled state? # ### SQL: # # SELECT count(*) FROM Lessons WHERE lesson_status_code = "Cancelled"; # ### End.
driving_school	### Complete SQL query only and with no explanation ### SQL tables followed by foreign key information: # # Addresses ( address_id, line_1_number_building, city, zip_postcode, state_province_county, country ) # Staff ( staff_id, staff_address_id, nickname, first_name, middle_name, last_name, date_of_birth, date_joined_staff, date_left_staff ) # Vehicles ( vehicle_id, vehicle_details ) # Customers ( customer_id, customer_address_id, customer_status_code, date_became_customer, date_of_birth, first_name, last_name, amount_outstanding, email_address, phone_number, cell_mobile_phone_number ) # Customer_Payments ( customer_id, datetime_payment, payment_method_code, amount_payment ) # Lessons ( lesson_id, customer_id, lesson_status_code, staff_id, vehicle_id, lesson_date, lesson_time, price ) # # Staff.staff_address_id can be joined with Addresses.address_id # Customers.customer_address_id can be joined with Addresses.address_id # Customer_Payments.customer_id can be joined with Customers.customer_id # Lessons.customer_id can be joined with Customers.customer_id # Lessons.staff_id can be joined with Staff.staff_id # Lessons.vehicle_id can be joined with Vehicles.vehicle_id # ### Question: # # How many lessons have been cancelled? # ### SQL: # # SELECT count(*) FROM Lessons WHERE lesson_status_code = "Cancelled"; # ### End.

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