鉴于这些关系,您如何查询以下内容: 预订至少退休金的游客(姓名和电子邮件)评分大于9 ,但没有预订任何三星级酒店 评级小于9 。
以下是否正确?
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
)
答案 0 :(得分:1)
我会使用聚合和having
:
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;
您的方法也有效,但在子查询中可能需要t.id
。