Question

我有一个表“事件”和一个表“Slots”，其中Slots.SlotID = Events.MainSlot OR Events.ExtraSlot。

我需要将每个时段的“参加者”数量相加（因为MainSlot和ExtraSlot - ExtraSlot是可选的）

表“事件”

ID------Name----------MainSlot-------ExtraSlot-------Attendees
1-------Event1--------1 -------------n/a-------------20
2-------Event2--------1 -------------n/a-------------20
3-------Event3--------2 -------------n/a-------------40
4-------Event4--------2 -------------3---------------20
5-------Event5--------3 -------------4---------------40
6-------Event6--------3 -------------4---------------20
7-------Event7--------3 -------------4---------------10

表“老虎机”

SlotID--- Slot
1-------- 9.00-9.30
2-------- 9.30-10
3-------- 10.30-10.30
4-------- 10.30-11

如果我按如下方式单独查询数据库：

         SELECT s.Slot, s.SlotID, ISNULL(SUM(e. Attendees), 0) AS Attendees1
           FROM Slots AS s 
LEFT OUTER JOIN Events AS e ON e.MainSlot = s.SlotID
       GROUP BY s.Slot, s.SlotID

......或：

         SELECT s.Slot, s.SlotID, ISNULL(SUM(x.Attendees), 0) AS Attendees2 
           FROM Slots AS s 
LEFT OUTER JOIN Events AS x ON x.ExtraSlot = s.SlotID 
       GROUP BY s.Slot, s.SlotID

我分别得到以下内容：

SlotID ------  Attendees1
1------------- 40
2------------- 60
3------------- 70
4------------- 0

SlotID ------- Attendees2
1------------- 0
2------------- 0
3------------- 20 *correct
4------------- 70

两种结果都是正确的。

但是如果我把两个查询放在一起就会出现问题，如下表所示

SELECT  s.Slot, s.SlotID, ISNULL(SUM(e.Attendees), 0) AS Attendees1,
        ISNULL(SUM(x. Attendees), 0) AS Attendees2
FROM Slots AS s LEFT OUTER JOIN
Events AS e ON e.MainSlot = s.SlotID LEFT OUTER JOIN
Events AS x ON x.ExtraSlot = s.SlotID
GROUP BY s.Slot, s.SlotID

SlotID------------- Attendees1---------- Attendees2
1-------------------40-------------------0
2-------------------60-------------------0
3-------------------70------------------60 *wrong
4-------------------0-------------------70

我做错了什么？谢谢你的帮助！

Answer 1

SELECT  s.SlotId,
        COALESCE(
        (
        SELECT  SUM(attendees)
        FROM    events ea
        WHERE   ea.MainSlot = s.SlotId
        ), 0) AS AttendeesAsMain,
        COALESCE(
        (
        SELECT  SUM(attendees)
        FROM    events ea
        WHERE   ea.ExtraSlot = s.SlotId
        ), 0) AS AttendeesAsExtra
FROM    Slots s

Answer 2

我的猜测是，这与你在该组中都有记录这一事实有关。请尝试以下（MSSQL Server）

SELECT s.Slot, s.SlotID, ISNULL(SUM(e.Attendees), 0) AS Attendees1,
ISNULL(SUM(CASE WHEN  x. Attendees IS NULL THEN 0 ELSE x.Attendees END), 0) AS Attendees2
FROM Slots AS s LEFT OUTER JOIN
Events AS e ON e.MainSlot = s.SlotID LEFT OUTER JOIN
Events AS x ON x.ExtraSlot = s.SlotID
GROUP BY s.Slot, s.SlotID

Answer 3

    SELECT a.SlotID, Attendees1, Attendees2 FROM
         (SELECT s.Slot, s.SlotID, ISNULL(SUM(e. Attendees), 0) AS Attendees1
           FROM Slots AS s 
         LEFT OUTER JOIN Events AS e ON e.MainSlot = s.SlotID
           GROUP BY s.Slot, s.SlotID) as a,

         (SELECT s.Slot, s.SlotID, ISNULL(SUM(x.Attendees), 0) AS Attendees2 
           FROM Slots AS s 
         LEFT OUTER JOIN Events AS x ON x.ExtraSlot = s.SlotID 
           GROUP BY s.Slot, s.SlotID) as b
    WHERE a.SlotID = b.SlotID

Answer 4

您加入了两次，因此您在返回时会得到重复的结果。基本上你得到这个：

slotID用于...... e.mainslot ...... x.extraslot ....... x.attendees ..... e.attendees

3 ........... 3 ............... 3 ................. 20 .............. 40个
3 ........... 3 ............... 3 ................. 20 ... ........... 20个
3 ........... 3 ............... 3 ................. 20 ... ........... 10

这是预期的行为，因为您将X连接到Slots JOIN E中的每个结果行。

两个SUM查询在一起

4 个答案: