1sf/bird-sql-train-with-schema huggingface dataset:
Database: bike_share_1
Tables: station, status, trip, weather
Columns (table index, column name): -1, *, 0, id, 0, name, 0, lat, 0, long, 0, dock_count, 0, city, 0, installation_date, 1, station_id, 1, bikes_available, 1, docks_available, 1, time, 2, id, 2, duration, 2, start_date, 2, start_station_name, 2, start_station_id, 2, end_date, 2, end_station_name, 2, end_station_id, 2, bike_id, 2, subscription_type, 2, zip_code, 3, date, 3, max_temperature_f, 3, mean_temperature_f, 3, min_temperature_f, 3, max_dew_point_f, 3, mean_dew_point_f, 3, min_dew_point_f, 3, max_humidity, 3, mean_humidity, 3, min_humidity, 3, max_sea_level_pressure_inches, 3, mean_sea_level_pressure_inches, 3, min_sea_level_pressure_inches, 3, max_visibility_miles, 3, mean_visibility_miles, 3, min_visibility_miles, 3, max_wind_Speed_mph, 3, mean_wind_speed_mph, 3, max_gust_speed_mph, 3, precipitation_inches, 3, cloud_cover, 3, events, 3, wind_dir_degrees, 3, zip_code
Question: Which station did the user who started at Market at 4th station ended their trip at the time of 12:45:00 PM and the date of 8/29/2013 and what is the location coordinates of the ending station?
Evidence: started at refers to start_station_name; start_station_name = 'Market at 4th'; location coordinates refers to (lat, long);
SQL: SELECT T1.name, T1.lat, T1.long FROM station AS T1 INNER JOIN trip AS T2 ON T2.end_station_name = T1.name WHERE T2.start_station_name = 'Market at 4th' AND T2.end_date = '8/29/2013 12:45'
Database: bike_share_1
Tables: station, status, trip, weather
Columns (table index, column name): -1, *, 0, id, 0, name, 0, lat, 0, long, 0, dock_count, 0, city, 0, installation_date, 1, station_id, 1, bikes_available, 1, docks_available, 1, time, 2, id, 2, duration, 2, start_date, 2, start_station_name, 2, start_station_id, 2, end_date, 2, end_station_name, 2, end_station_id, 2, bike_id, 2, subscription_type, 2, zip_code, 3, date, 3, max_temperature_f, 3, mean_temperature_f, 3, min_temperature_f, 3, max_dew_point_f, 3, mean_dew_point_f, 3, min_dew_point_f, 3, max_humidity, 3, mean_humidity, 3, min_humidity, 3, max_sea_level_pressure_inches, 3, mean_sea_level_pressure_inches, 3, min_sea_level_pressure_inches, 3, max_visibility_miles, 3, mean_visibility_miles, 3, min_visibility_miles, 3, max_wind_Speed_mph, 3, mean_wind_speed_mph, 3, max_gust_speed_mph, 3, precipitation_inches, 3, cloud_cover, 3, events, 3, wind_dir_degrees, 3, zip_code
Question: How many subscribers have ended their trip at MLK Library and how many docks does that station have?
Evidence: subscribers refers to subscription_type = 'subscribers'; ended their trip at refers to end_station_name; end_station_name = 'MLK Library'; number of docks a station have refers to dock_count;
SQL: SELECT COUNT(T1.id), T2.dock_count FROM trip AS T1 INNER JOIN station AS T2 ON T2.name = T1.start_station_name WHERE T1.end_station_name = 'MLK Library' AND T1.subscription_type = 'Subscriber' AND T2.dock_count = 19
Database: bike_share_1
Tables: station, status, trip, weather
Columns (table index, column name): -1, *, 0, id, 0, name, 0, lat, 0, long, 0, dock_count, 0, city, 0, installation_date, 1, station_id, 1, bikes_available, 1, docks_available, 1, time, 2, id, 2, duration, 2, start_date, 2, start_station_name, 2, start_station_id, 2, end_date, 2, end_station_name, 2, end_station_id, 2, bike_id, 2, subscription_type, 2, zip_code, 3, date, 3, max_temperature_f, 3, mean_temperature_f, 3, min_temperature_f, 3, max_dew_point_f, 3, mean_dew_point_f, 3, min_dew_point_f, 3, max_humidity, 3, mean_humidity, 3, min_humidity, 3, max_sea_level_pressure_inches, 3, mean_sea_level_pressure_inches, 3, min_sea_level_pressure_inches, 3, max_visibility_miles, 3, mean_visibility_miles, 3, min_visibility_miles, 3, max_wind_Speed_mph, 3, mean_wind_speed_mph, 3, max_gust_speed_mph, 3, precipitation_inches, 3, cloud_cover, 3, events, 3, wind_dir_degrees, 3, zip_code
Question: What is the average coldest temperature for the zip code of 94301 and what stations are within the zip code? Include the latitude and longitude as well.
Evidence: coldest temperature refers to min_temperature_f; average coldest temperature refers = AVG(min_temperature_f); stations refers to name; latitude refers to lat; longitude refers to long;
SQL: SELECT AVG(T3.min_temperature_f), T1.long, T1.lat FROM station AS T1 INNER JOIN trip AS T2 ON T2.start_station_name = T1.name INNER JOIN weather AS T3 ON T3.zip_code = T2.zip_code WHERE T3.zip_code = 94301
Database: bike_share_1
Tables: station, status, trip, weather
Columns (table index, column name): -1, *, 0, id, 0, name, 0, lat, 0, long, 0, dock_count, 0, city, 0, installation_date, 1, station_id, 1, bikes_available, 1, docks_available, 1, time, 2, id, 2, duration, 2, start_date, 2, start_station_name, 2, start_station_id, 2, end_date, 2, end_station_name, 2, end_station_id, 2, bike_id, 2, subscription_type, 2, zip_code, 3, date, 3, max_temperature_f, 3, mean_temperature_f, 3, min_temperature_f, 3, max_dew_point_f, 3, mean_dew_point_f, 3, min_dew_point_f, 3, max_humidity, 3, mean_humidity, 3, min_humidity, 3, max_sea_level_pressure_inches, 3, mean_sea_level_pressure_inches, 3, min_sea_level_pressure_inches, 3, max_visibility_miles, 3, mean_visibility_miles, 3, min_visibility_miles, 3, max_wind_Speed_mph, 3, mean_wind_speed_mph, 3, max_gust_speed_mph, 3, precipitation_inches, 3, cloud_cover, 3, events, 3, wind_dir_degrees, 3, zip_code
Question: Calculate the average duration travelled by subscribers that both started and ended their trip in Mountain View City Hall and indicate the date when the station was first installed.
Evidence: average duration = DIVIDE(SUM(duration), COUNT(id)); subscribers refers to subscription_type = 'subscriptions'; started and ended their trip at Mountain View City Hall refers to start_station_name = 'Mountain View City Hall' and end_station_name = 'Mountain View City Hall'; when the station was first installed refers to installation_date;
SQL: SELECT AVG(T1.duration), T2.installation_date FROM trip AS T1 INNER JOIN station AS T2 ON T2.name = T1.start_station_name WHERE T1.start_station_name = 'Mountain View City Hall' AND T1.subscription_type = 'Subscriber' AND T1.end_station_name = 'Mountain View City Hall'
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: What is the description of the film ACADEMY DINOSAUR?
Evidence: "ACADEMY DINOSAUR" is the title of film
SQL: SELECT description FROM film WHERE title = 'ACADEMY DINOSAUR'
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: How many films have a rental duration of over 6 days?
Evidence: rental duration of over 6 days refers to rental_duration > 6
SQL: SELECT COUNT(film_id) FROM film WHERE rental_duration > 6
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Please list the titles of the films that are released in 2006 and have a rental rate of $2.99.
Evidence: released in 2006 refers to release_year = 2006; rental rate of $2.99 refers to rental_rate = 2.99
SQL: SELECT title FROM film WHERE release_year = 2006 AND rental_rate = 2.99
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Which film has the longest duration of film screening? Please give its title.
Evidence: longest duration of film refers to Max(length)
SQL: SELECT title FROM film ORDER BY length DESC LIMIT 1
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Which film has a higher replacement cost, ACE GOLDFINGER or ACADEMY DINOSAUR?
Evidence: higher replacement cost refers to Max(replacement_cost); 'ACE GOLDFIINGER' and 'ACADEMY DINOSAUR' are both the title of film
SQL: SELECT title FROM film WHERE title IN ('ACE GOLDFINGER', 'ACADEMY DINOSAUR') ORDER BY replacement_cost DESC LIMIT 1
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Among the films that are released in 2006, how many of them are rated Adults Only in the Motion Picture Association Film Rating?
Evidence: released in 2006 refers to release_year = 2006; rated Adults Only refers to rating = 'NC-17'
SQL: SELECT COUNT(film_id) FROM film WHERE rating = 'NC-17' AND release_year = 2006
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: How many films with the rental rate of $2.99 have the special feature of "Deleted Scenes"?
Evidence: rental rate of $2.99 refers to rental_rate = 2.99; film refers to title
SQL: SELECT COUNT(film_id) FROM film WHERE rental_rate = 2.99 AND special_features = 'Deleted Scenes'
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Please list the titles of all the films that have more than 2 special features.
Evidence: more than 2 special features refers to Count(special_features) > 2
SQL: SELECT title FROM ( SELECT title, COUNT(special_features) AS num FROM film GROUP BY title ) AS T ORDER BY T.num > 2
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: What is the email address of the staff Jon Stephens?
SQL: SELECT email FROM staff WHERE first_name = 'Jon' AND last_name = 'Stephens'
Database: movie_3
Tables: film_text, actor, address, category, city, country, customer, film, film_actor, film_category, inventory, language, payment, rental, staff, store
Columns (table index, column name): -1, *, 0, film_id, 0, title, 0, description, 1, actor_id, 1, first_name, 1, last_name, 1, last_update, 2, address_id, 2, address, 2, address2, 2, district, 2, city_id, 2, postal_code, 2, phone, 2, last_update, 3, category_id, 3, name, 3, last_update, 4, city_id, 4, city, 4, country_id, 4, last_update, 5, country_id, 5, country, 5, last_update, 6, customer_id, 6, store_id, 6, first_name, 6, last_name, 6, email, 6, address_id, 6, active, 6, create_date, 6, last_update, 7, film_id, 7, title, 7, description, 7, release_year, 7, language_id, 7, original_language_id, 7, rental_duration, 7, rental_rate, 7, length, 7, replacement_cost, 7, rating, 7, special_features, 7, last_update, 8, actor_id, 8, film_id, 8, last_update, 9, film_id, 9, category_id, 9, last_update, 10, inventory_id, 10, film_id, 10, store_id, 10, last_update, 11, language_id, 11, name, 11, last_update, 12, payment_id, 12, customer_id, 12, staff_id, 12, rental_id, 12, amount, 12, payment_date, 12, last_update, 13, rental_id, 13, rental_date, 13, inventory_id, 13, customer_id, 13, return_date, 13, staff_id, 13, last_update, 14, staff_id, 14, first_name, 14, last_name, 14, address_id, 14, picture, 14, email, 14, store_id, 14, active, 14, username, 14, password, 14, last_update, 15, store_id, 15, manager_staff_id, 15, address_id, 15, last_update
Question: Please give the full names of all the active staff.
Evidence: full name refers to first_name, last_name; active staff refers to active = 1
SQL: SELECT first_name, last_name FROM staff WHERE active = 1