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 three types of film along with their IDs and the latest update.
Evidence: types of film refers to the name of a category; IDs refers to category_id; latest update refers to last_update.
SQL: SELECT DISTINCT name, category_id, last_update FROM category LIMIT 3
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 full names of any three inactive customers.
Evidence: full name refers to first_name, last_name; inactive customers refers to active = 0
SQL: SELECT first_name, last_name FROM customer WHERE active = 0 LIMIT 3
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 rental price per day for Airplane Sierra?
Evidence: rental price per day refers to DIVIDE(rental_price, rental_duration); 'Airplane Sierra' is a title of a film
SQL: SELECT rental_rate / rental_duration AS result FROM film WHERE title = 'AIRPLANE SIERRA'
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: Where is store number 2 located?
Evidence: store number 2 refers to store_id = 2; where is a store located refers to address, address2, district
SQL: SELECT T1.address, T1.address2, T1.district FROM address AS T1 INNER JOIN store AS T2 ON T1.address_id = T2.address_id WHERE T2.store_id = 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: Which city does the address 1623 Kingstown Drive belong to?
SQL: SELECT T1.city FROM city AS T1 INNER JOIN address AS T2 ON T2.city_id = T1.city_id WHERE T2.address = '1623 Kingstown Drive'
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 name three cities that belong to Algeria.
Evidence: Algeria is a country
SQL: SELECT T2.city FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id WHERE T1.country = 'Algeria'
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 category of the film Agent Truman?
Evidence: 'Agent Truman' is a title of a film; category refers to name
SQL: SELECT T3.name FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T1.title = 'AGENT TRUMAN'
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 any three action films.
Evidence: action is a name of category
SQL: SELECT T1.title FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Action' LIMIT 3
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 difference between the number of children's films and action films?
Evidence: 'children' AND 'action' are names of a category; Calculation = SUBTRACT(AVG('children'), AVG('action'))
SQL: SELECT SUM(IIF(T2.name = 'Children', 1, 0)) - SUM(IIF(T2.name = 'Action', 1, 0)) AS diff 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: Which district does Maria Miller live in?
Evidence: 'Maria Miller' is a name of a customer; full name refers to first_name, last_name
SQL: SELECT T2.district FROM customer AS T1 INNER JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.first_name = 'Maria' AND T1.last_name = 'Miller'
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 customer that is active and lives at 1795 Santiago de Compostela Way, Texas?
Evidence: active refers to active = 1; '1795 Santiago de Compostela Way' is an address; Texas is a district; who refers to first_name, last_name
SQL: SELECT T1.first_name, T1.last_name FROM customer AS T1 INNER JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T2.address = '1795 Santiago de Compostela Way' AND T1.active = 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: How many English films have a duration of over 50 minutes and the cost of replacement are under 10.99?
Evidence: English is a name of a language; duration of over 50 minutes refers to length > 50; cost of replacement are under 10.99 refers to replacement_cost < 10.99
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.length > 50 AND T1.replacement_cost < 10.99
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: Who are the actors that act in the ACADEMY DINOSAUR film?
Evidence: Who are the actors refers to full name; full name refers to first_name, last_name; 'ACADEMY DINOSAUR' is a title of a film
SQL: SELECT T1.first_name, T1.last_name 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 T3.title = 'ACADEMY DINOSAUR'