human
Schema:
actor (actor_id, first_name, last_name, last_update)
address (address_id, address, address2, district, city_id, postal_code, phone, last_update)
category (category_id, name, last_update)
city (city_id, city, country_id, last_update)
country (country_id, country, last_update)
customer (customer_id, store_id, first_name, last_name, email, address_id, active, create_date, last_update)
film (film_id, title, description, release_year, language_id, original_language_id, rental_duration, rental_rate, length, replacement_cost, rating, special_features, last_update)
film_actor (actor_id, film_id, last_update)
film_category (film_id, category_id, last_update)
film_text (film_id, title, description)
inventory (inventory_id, film_id, store_id, last_update)
language (language_id, name, last_update)
payment (payment_id, customer_id, staff_id, rental_id, amount, payment_date, last_update)
rental (rental_id, rental_date, inventory_id, customer_id, return_date, staff_id, last_update)
staff (staff_id, first_name, last_name, address_id, picture, email, store_id, active, username, password, last_update)
store (store_id, manager_staff_id, address_id, last_update)
Please answer with SQL.
assistant
Sure, I understand.
human
Tell me the email of customer with first name PATRICIA.
assistant
SELECT email FROM customer WHERE first_name = 'PATRICIA';
human
Tell me all the information about the payments they made.
assistant
SELECT * FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id WHERE T1.first_name = 'PATRICIA';
human
tell me which customer, who has made at least one payment, has spent the least money. List his or her first name, last name, and the id.
assistant
SELECT T1.first_name , T1.last_name , T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM( amount ) ASC LIMIT 1 ;