Group By子句是否需要加入？

Question

鑑於這種模式：

Hotel(hotelNo, hotelName, city)      #hotelNo is PK 
Room (roomNo, hotelNo, type, price)     #(roomNo, hotelNo) is PK 
Booking (hotelNo, guestNo, dateFrom, dateTo, roomNo) #(hotelNo,guestNo, dateFrom) is PK 
Guest(guestNo, GuestName, GuestAddress)    #guestNo is PK 

問：什麼是從空閒的客房收入損失在今天每家酒店？

看着我結束了一個非常不同的答案（更復雜，但解決方案基本上我發現all rooms in hotels並計算Difference集合運算符對all rooms which are booked today，然後計算出的價格與group by子句剩下的房間 - 我米甚至不知道它會工作，但在紙上看起來有效）。但是，看看示例解決方案：

SELECT hotelNo, SUM(price) FROM Room r 
WHERE roomNo NOT IN 
    (SELECT roomNo FROM Booking b, Hotel h 
    WHERE (dateFrom <= CURRENT_DATE AND 
     dateTo >= CURRENT_DATE) AND 
     b.hotelNo = h.hotelNo) 
GROUP BY hotelNo; 

爲什麼我們需要加入預訂和酒店。我看到我們基於b.hotelNo=h.hotelNo基於b.hotelNo=h.hotelNo加入這兩個關係，但查看架構，Booking提供了RoomNo（已預訂）的詳細信息，並提供關於哪個酒店和日期的信息......當然，這是我們需要處理的所有信息一個Group By子句並實現我們想要的結果？還是我誤會了？

Answer 1

你不需要hotel表。假設所有預訂實際上都是酒店，booking中的外鍵值可用於比較。

這應該做工精細，即使兩個HotelNo和RoomNo需要確定一個預訂和客房：

SELECT hotelNo, SUM(price) 
FROM Room r 
WHERE r.roomNo NOT IN (SELECT b.roomNo 
         FROM Booking b 
         WHERE (dateFrom <= CURRENT_DATE AND dateTo >= CURRENT_DATE) AND 
          b.hotelNo = r.hotelNo 
        ) 
GROUP BY hotelNo; 

然而，尷尬的分裂這樣的配對關係。使用not exists代替：

SELECT hotelNo, SUM(price) 
FROM Room r 
WHERE NOT EXISTS (SELECT 1 
        FROM Booking b 
        WHERE (dateFrom <= CURRENT_DATE AND dateTo >= CURRENT_DATE) AND 
         b.hotelNo = r.hotelNo and 
         b.roomNo = r.roomNo 
       ) 
GROUP BY hotelNo; 

當然，你可以這樣寫爲join，但即使有join，不需要hotel表：

SELECT r.hotelNo, SUM(r.price) 
FROM Room r left join 
    Booking b 
    ON b.dateFrom <= CURRENT_DATE AND 
     b.dateTo >= CURRENT_DATE and 
     b.roomNo = r.roomNo and b.hotelNo = r.hotelNo 
WHERE b.roomNo is null  
GROUP BY r.hotelNo; 

Answer 2

如果我理解正確的話，這更是我會寫的查詢類型：

SELECT hotelNo, SUM(price) FROM Room r 
left join 
    (SELECT roomNo , hotelNo 
    FROM Booking b 
    join Hotel h ON b.hotelNo = h.hotelNo 
    WHERE dateFrom <= CURRENT_DATE AND 
     dateTo >= CURRENT_DATE) a 
     ON a. roomNo = r.roomNo and a.hotelNo= r.hotelNo 
WHERE a.roomNo is null  
GROUP BY hotelNo;