1sf/bird-sql-train-with-schema huggingface dataset:
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 any two films that Penelope Guiness acted in.
Evidence: film refers to title of the film; 'Penelope Guiness' is a full name of an actor; full name refers to first_name, last_name
SQL: SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'Penelope' AND T1.last_name = 'Guiness' LIMIT 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 percentage of documentary films?
Evidence: documentary' is a name of a category; calculation = DIVIDE(SUM(name = 'Documentary'), COUNT(film_id)) * 100
SQL: SELECT CAST(SUM(IIF(T2.name = 'Documentary', 1, 0)) AS REAL) * 100 / COUNT(T1.film_id) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id
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 in English are for adults only?
Evidence: English is a name of a language; for adults only refers to rating = 'NC-17'
SQL: SELECT COUNT(T1.film_id) FROM film AS T1 INNER JOIN language AS T2 ON T1.language_id = T2.language_id WHERE T2.name = 'English' AND T1.rating = 'NC-17'
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?
Evidence: film refers to the title; the longest duration refers to MAX(length)
SQL: SELECT title FROM film WHERE length = ( SELECT MAX(length) FROM film )
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 of the actors are named "Dan"?
Evidence: 'Dan' is a first_name of an actor
SQL: SELECT COUNT(actor_id) FROM actor WHERE first_name = 'Dan'
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 most common first name among the customers?
Evidence: the most common first name refers to MAX(COUNT(first_name))
SQL: SELECT first_name FROM customer GROUP BY first_name ORDER BY COUNT(first_name) 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: What are the ratings of the film featuring behind the scenes?
Evidence: film featuring behind the scenes refers to special_features = 'Behind the Scenes'
SQL: SELECT rating FROM film WHERE special_features LIKE '%Behind the 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: What is the largest number of films rented per customer?
Evidence: the largest number of films refers to MAX(rental_id)
SQL: SELECT COUNT(rental_id) FROM rental GROUP BY customer_id ORDER BY COUNT(rental_id) 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: List all the films with the word "Lacklusture" in their description.
Evidence: films refers to title
SQL: SELECT title FROM film_text WHERE description LIKE '%Lacklusture%'
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 did a customer named Francis Sikes rent?
Evidence: 'Francis Sikes' is a full name of a customer; full name refers to first_name, last_name;
SQL: SELECT COUNT(T1.customer_id) FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id WHERE T1.first_name = 'FRANCIS' AND T1.last_name = 'SIKES'
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: Who is the manager of the store with the largest collection of films?
Evidence: Who refers to first_name, last_name; the largest collection of films refers to MAX(film_id)
SQL: SELECT T.first_name, T.last_name FROM ( SELECT T3.first_name, T3.last_name, COUNT(T1.film_id) AS num FROM inventory AS T1 INNER JOIN store AS T2 ON T1.store_id = T2.store_id INNER JOIN staff AS T3 ON T2.manager_staff_id = T3.staff_id GROUP BY T3.first_name, T3.last_name ) AS T ORDER BY T.num 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: What are the addresses of the inactive customers?
Evidence: inactive customers refers to active = 0;
SQL: SELECT T2.address FROM customer AS T1 INNER JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.active = 0
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 category is the most common?
Evidence: most common category refers to MAX(COUNT(category.name))
SQL: SELECT T.name FROM ( SELECT T2.name, COUNT(T2.name) AS num FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T2.name ) AS T ORDER BY T.num DESC LIMIT 1