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: Provide the cast for the film "Jason trap".
Evidence: 'Jason trap' is a title of a film; cast means actor; actor refers to first_name, last_name
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 = 'JASON TRAP'
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 with the largest payment for rental films?
Evidence: Who refers to first_name, last_name; the largest payment for rental refers to MAX(SUM(amount))
SQL: SELECT T.first_name, T.last_name FROM ( SELECT T1.first_name, T1.last_name, SUM(T2.amount) AS num FROM customer AS T1 INNER JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.first_name, T1.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: List the top 5 most-rented films.
Evidence: film refers to title; most rented refers to MAX(inventory_id)
SQL: SELECT T.title FROM ( SELECT T3.title, COUNT(T2.inventory_id) AS num FROM rental AS T1 INNER JOIN inventory AS T2 ON T1.inventory_id = T2.inventory_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id GROUP BY T3.title ) AS T ORDER BY T.num DESC LIMIT 5
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 country does Sasebo belong to?
Evidence: 'Sasebo' is a city
SQL: SELECT T1.country FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id WHERE T2.city = 'Sasebo'
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 for the stores?
SQL: SELECT T2.address FROM store AS T1 INNER JOIN address AS T2 ON T1.address_id = T2.address_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: List all the animation titles.
Evidence: 'animation' is a name of a category
SQL: SELECT T3.title AS per FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id INNER JOIN film AS T3 ON T1.film_id = T3.film_id WHERE T2.name = 'Animation'
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 city with the most customers?
Evidence: the most customers refers to MAX(COUNT(customer_id))
SQL: SELECT T.city FROM ( SELECT T1.city, COUNT(T3.customer_id) AS num FROM city AS T1 INNER JOIN address AS T2 ON T2.city_id = T1.city_id INNER JOIN customer AS T3 ON T2.address_id = T3.address_id GROUP BY T1.city ) 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: Which actor acted in the most films?
Evidence: actor refers to first_name, last_name; the most film refers to MAX(SUM(film_id))
SQL: SELECT T.first_name, T.last_name FROM ( SELECT T2.first_name, T2.last_name, SUM(T1.film_id) AS num FROM film_actor AS T1 INNER JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.first_name, T2.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 percentage of films are horror films?
Evidence: horror' is a name of a category; calculation = DIVIDE(SUM(name = 'Horror'), COUNT(film_id)) * 100
SQL: SELECT CAST(SUM(IIF(T2.name = 'Horror', 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 INNER JOIN film AS T3 ON T1.film_id = T3.film_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: Please indicate the full name of actor id 5.
Evidence: full name refers to first_name, last_name
SQL: SELECT first_name, last_name FROM actor WHERE actor_id = 5
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 id movies have category id 11?
Evidence: id movies refers to film_id
SQL: SELECT COUNT(film_id) FROM film_category WHERE category_id = 11
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 does BABY HALL film belong to?
Evidence: category refers to name; BABY HALL film refers to title = 'BABY HALL'
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 T3.category_id = T2.category_id WHERE T1.title = 'BABY 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: Give the full name of the actor with the highest rental rate.
Evidence: full name refers to first_name, last_name; the highest rental rate refers to max(rental_rate)
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 T3.film_id = T2.film_id ORDER BY T3.rental_rate DESC LIMIT 1