Rentals: Подтвержденные бронирования

Easy

Войдите, чтобы сохранялся прогресс.

Подсчитайте количество бронирований, где status = confirmed. Ожидаемая колонка: confirmed_cnt

Структура таблиц

PRAGMA foreign_keys = ON;
  CREATE TABLE hosts (id INTEGER PRIMARY KEY, host_name TEXT NOT NULL, country TEXT NOT NULL);
  CREATE TABLE listings (id INTEGER PRIMARY KEY, host_id INTEGER NOT NULL, title TEXT NOT NULL, city TEXT NOT NULL, nightly_price INTEGER NOT NULL, max_guests INTEGER NOT NULL,
    FOREIGN KEY(host_id) REFERENCES hosts(id));
  CREATE TABLE bookings (id INTEGER PRIMARY KEY, listing_id INTEGER NOT NULL, guest_name TEXT NOT NULL, checkin TEXT NOT NULL, nights INTEGER NOT NULL, guests INTEGER NOT NULL, status TEXT NOT NULL,
    FOREIGN KEY(listing_id) REFERENCES listings(id));
  CREATE TABLE reviews (id INTEGER PRIMARY KEY, booking_id INTEGER NOT NULL, rating INTEGER NOT NULL, comment TEXT,
    FOREIGN KEY(booking_id) REFERENCES bookings(id));

Пример данных

bookings

id	listing_id	guest_name	checkin	nights	guests	status
1	2	Guest 1	2026-01-02	2	2	confirmed
2	3	Guest 2	2026-01-03	3	3	confirmed
3	4	Guest 3	2026-01-04	4	4	confirmed
4	5	Guest 4	2026-01-05	5	5	confirmed
5	6	Guest 5	2026-01-06	6	1	confirmed

Показано строк: 5 (LIMIT 5)

hosts

id	host_name	country
1	Host 1	DE
2	Host 2	FR
3	Host 3	ES
4	Host 4	NL
5	Host 5	DE

Показано строк: 5 (LIMIT 5)

listings

id	host_id	title	city	nightly_price	max_guests
1	2	Listing 1	Utrecht	70	2
2	3	Listing 2	Rotterdam	80	3
3	4	Listing 3	Eindhoven	90	4
4	5	Listing 4	Amsterdam	100	5
5	6	Listing 5	Utrecht	110	1

Показано строк: 5 (LIMIT 5)

reviews

id	booking_id	rating	comment
1	2	2	good
2	3	3	great
3	4	4	ok
4	5	5	good
5	6	1	great

Показано строк: 5 (LIMIT 5)

Ваш SQL

Результат

Результат пуст (0 строк).