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: How many films have a rental rate of 0.99?
SQL: SELECT COUNT(film_id) FROM film WHERE rental_rate = 0.99
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: Among the customers with customer ID of 100 and below, how many of them have Thomas as their last name?
Evidence: customer ID of 100 and below refers to customer_id < 100
SQL: SELECT COUNT(customer_id) FROM customer WHERE last_name = 'Thomas' AND customer_id < 100
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 actor's last name that starred the film with the description of "A Thoughtful Drama of a Composer And a Feminist who must Meet a Secret Agent in The Canadian Rockies".
SQL: SELECT 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.description = 'A Thoughtful Drama of a Composer And a Feminist who must Meet a Secret Agent in The Canadian Rockies'
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 title of the film starred by Liza Bergman with the highest replacement cost.
Evidence: Liza Bergman' is a full name; full name refers to first_name, last_name; highest replacement cost refers to MAX(replacement_cost)
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 = 'Liza' AND T1.last_name = 'Bergman' 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 films with store ID of 2, list the title of films with the highest rental rate.
Evidence: highest rental rate refers to MAX(rental_rate)
SQL: SELECT T1.title FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id WHERE T2.store_id = 2 ORDER BY rental_rate 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 starred by Angelina Astaire, what is the title of the film with a replacement cost of 27.99?
Evidence: Angelina Astaire' 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 = 'Angelina' AND T1.last_name = 'Astaire' AND T3.replacement_cost = 27.99
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: List the inventory ID of the film titled "African Egg".
Evidence: 'African Egg' is a title of a film
SQL: SELECT T2.inventory_id FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id WHERE T1.title = 'African Egg'
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: In films with a length duration of 113 minutes, how many of the films are starred by Kirk Jovovich?
Evidence: length duration of 113 minutes refers to length = 113; 'Kirk Jovovich' is a full name of an actor; full name refers to first_name, last_name
SQL: SELECT COUNT(T1.actor_id) 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.length = 113 AND T1.first_name = 'Kirk' AND T1.last_name = 'Jovovich'
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: In the film with an inventory ID between 20 to 60, how many of the films have a G rating?
Evidence: G rating refers to rating = 'G'
SQL: SELECT COUNT(T1.film_id) FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id WHERE T2.inventory_id BETWEEN 20 AND 60 AND T1.rating = 'G'
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 films with a rental rate of 4.99, what is the total number of films starred by Bob Fawcett?
Evidence: Bob Fawcett' is a full name of an actor; full name refers to first_name, last_name
SQL: SELECT COUNT(T1.actor_id) 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.rental_rate = 4.99 AND T1.first_name = 'Bob' AND T1.last_name = 'Fawcett'
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 inventory ID of the films starred by Russell Close with a duration between 110 to 150 minutes?
Evidence: 'Russell Close' is a full name of an actor; full name refers to first_name, last_name; duration between 110 to 150 minutes refers to length BETWEEN 110 AND 150
SQL: SELECT T4.inventory_id 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 INNER JOIN inventory AS T4 ON T3.film_id = T4.film_id WHERE T3.length BETWEEN 110 AND 150 AND T1.first_name = 'Russell' AND T1.last_name = 'Close'
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 store and inventory ID of the film with the longest duration?
Evidence: the longest duration refers to MAX(length)
SQL: SELECT T2.store_id, T2.inventory_id FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id ORDER BY T1.length DESC LIMIT 1