以下のMy SQL Queryは、SQLコマンドがnumber_of_roomsをNULLで減算し、NULLをColumnに出力する、0の部屋をとった場合を除いて、RoomsAvailableフィールドを正しく返しています。私は数多くのISNULLバリエーションを試して、それがうまく動作していないことを発見しました。誰が私はこれをやっているべきか知っていますか?クエリ内のクエリでISNULLを使用するには?
SQL:
SELECT
Hotel_2.hotel_code,
Hotel_2.hotel_country,
Room_type_rates_2.room_type_code,
Room_type_rates_2.number_of_rooms,
Types_2.room_type,
Room_type_rates_2.rates,
Room_type_rates_2.number_of_rooms -
(SELECT
DISTINCT (SELECT
COUNT(dbo.Hotel.hotel_code) AS RoomsTake
FROM
dbo.Hotel
INNER JOIN dbo.Hotel_Reservation
ON dbo.Hotel.hotel_code = dbo.Hotel_Reservation.hotel_code
INNER JOIN dbo.Room_type_rates
ON dbo.Hotel.hotel_code = dbo.Room_type_rates.hotel_code
INNER JOIN dbo.Types
ON dbo.Hotel_Reservation.room_type_code = dbo.Types.room_type_code
AND dbo.Room_type_rates.room_type_code = dbo.Types.room_type_code
WHERE
(dbo.Room_type_rates.room_type_code = Room_type_rates_1.room_type_code)
AND (dbo.Hotel.hotel_code = Hotel_1.hotel_code)
AND (dbo.Hotel_Reservation.checkin_date >= Hotel_Reservation_1.checkin_date)
AND (dbo.Hotel_Reservation.checkout_date <= Hotel_Reservation_1.checkout_date)
) AS RoomsTaken
FROM
dbo.Hotel AS Hotel_1
INNER JOIN dbo.Hotel_Reservation AS Hotel_Reservation_1
ON Hotel_1.hotel_code = Hotel_Reservation_1.hotel_code
INNER JOIN dbo.Room_type_rates AS Room_type_rates_1
ON Hotel_1.hotel_code = Room_type_rates_1.hotel_code
INNER JOIN dbo.Types AS Types_1
ON Hotel_Reservation_1.room_type_code = Types_1.room_type_code
AND Room_type_rates_1.room_type_code = Types_1.room_type_code
WHERE
(Hotel_Reservation_1.checkin_date >= '11/19/2011')
AND (Hotel_Reservation_1.checkout_date <= '12/01/2011')
AND (Hotel_1.hotel_country = 'Adelaide')
AND (Types_1.room_type_code = Types_2.room_type_code)
) AS RoomsAvailable
FROM
dbo.Hotel AS Hotel_2
INNER JOIN dbo.Room_type_rates AS Room_type_rates_2
ON Hotel_2.hotel_code = Room_type_rates_2.hotel_code
INNER JOIN dbo.Types AS Types_2
ON Room_type_rates_2.room_type_code = Types_2.room_type_code
電流出力:あなたは、これが
...ISNULL(COUNT(dbo.Hotel.*),0)...
それとも、全くNULL
を取得している問題ですが?:
ADL20 Adelaide CPL 6 Couple Suite 514.0000 3
ADL20 Adelaide FYU 3 Family Suite 533.0000 2
ADL20 Adelaide KNG 2 King's Bedroom 556.0000 NULL
なぜSELECT DISTNCTにSELECT COUNTがありますか?クエリで複数のレコードが返された場合は、おそらくエラーが発生し、計算が機能しなくなります。 – vol7ron