SQLクエリが正しいかどうか

Question

、次のクエリを実行できる方法：定格を超える9ある少なくとも年金を予約 観光客（名前と電子メール）が、 以下の3つ星ホテルを予約していない。

以下は正しいですか？

SELECT Tourists.name, Tourists.email 
FROM Tourists 
WHERE EXISTS (
    SELECT id FROM Bookings 
    INNER JOIN Tourists ON Bookings.touristId=Tourists.id 
    INNER JOIN AccomodationEstablishments ON Bookings.accEstId=AccomodationEstablishments.id 
    INNER JOIN AccomodationTypes ON AccomodationEstablishments.accType=AccomodationTypes.id 
    WHERE AccomodationTypes.name = 'Pension' AND 
     AccomodationEstablishments.rating > 9 
) AND NOT EXISTS (
    SELECT id FROM Bookings 
    INNER JOIN Tourists ON Bookings.touristId=Tourists.id 
    INNER JOIN AccomodationEstablishments ON Bookings.accEstId=AccomodationEstablishments.id 
    INNER JOIN AccomodationTypes ON AccomodationEstablishments.accType=AccomodationTypes.id 
    WHERE AccomodationTypes.name = 'Hotel' AND 
     AccomodationEstablishments.noOfStars = 3 AND 
     AccomodationEstablishments.rating < 9 
) 

Answer 1

私はこの使用して集約し、havingだろう：

SELECT t.name, t.email 
FROM Bookings b INNER JOIN 
    Tourists t 
    ON b.touristId = t.id INNER JOIN 
    AccomodationEstablishments ae 
    ON b.accEstId = ae.id INNER JOIN 
    AccomodationTypes a 
    ON ae.accType = a.id 
GROUP BY t.name, t.email 
HAVING SUM(CASE WHEN a.name = 'Pension' AND ae.rating > 9 THEN 1 ELSE 0 END) > 0 AND 
     SUM(a.name = 'Hotel' AND ae.noOfStars = 3 AND ae.rating < 9 THEN 1 ELSE 0 END)= 0; 

をあなたの方法も動作しますが、あなたはおそらくサブクエリでt.idを必要としています。