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
What are the districts of the addresses?
assistant
SELECT district FROM address;
human
How about the district which has at least two addresses?
assistant
SELECT district FROM address GROUP BY district HAVING COUNT( * ) > = 2 ;