SQL查询第二个求和列平方结果答案

【问题标题】：SQL Query Second sum column squaring resultsSQL查询第二个求和列平方结果
【发布时间】：2014-05-15 09:24:47
【问题描述】：

如果我分别运行 2 个查询，我会得到这样的结果..

 select A.ACTNUMST, sum(B.EXTDCOST) as [IV Total]
 from GL00105 A
 INNER JOIN SEE30303 B on A.ACTINDX = B.IVIVINDX
 group by A.ACTNUMST

 Results -
 Account No      IV Total
 2101-00-137     2033.60
 4101-00-137     83765.86
 6101-00-137     301984.23

第二次查询

 select A.ACTNUMST as [Account No], SUM(C.PERDBLNC) as [GL Total]
 from GL00105 A
 LEFT JOIN GL10110 C on A.ACTINDX = C.ACTINDX
 group by A.ACTNUMST

 Results -
 Account No      GL Total
 2101-00-137     2033.60
 4101-00-137     83765.86
 6101-00-137     302656.23

我希望能够将两个结果结合在一起进行比较，但我相信它会重复 GL 总数中每一行的总和，然后再次对其求和，它会出现像 -

这样的大数字

 select A.ACTNUMST as [Account No], sum(B.EXTDCOST) as [IV Total], SUM(C.PERDBLNC) as [GL Total]
 from GL00105 A
 INNER JOIN SEE30303 B on A.ACTINDX = B.IVIVINDX
 LEFT JOIN GL10110 C on A.ACTINDX = C.ACTINDX
 group by A.ACTNUMST

 Results -
 Account No      IV Total        GL Total
 2101-00-137     2033.60         14235.20
 4101-00-137     83765.86        116350696.20
 6101-00-137     301984.23       1612897825.84

应该是什么时候

 Results -
 Account No      IV Total        GL Total
 2101-00-137     2033.60         2033.60
 4101-00-137     83765.86        83765.86
 6101-00-137     301984.23       302656.23

请告知如何使用 sum 函数得到正确的结果。

【问题讨论】：

标签： sql sql-server-2008 group-by sum

【解决方案1】：

你想要这样的东西吗？如果这不是预期的，请检查并发表评论。

还请纠正任何语法错误。

SELECT T1.[Account No], T1.[IV Total], T2.[GL Total] FROM
(
    select A.ACTNUMST as [Account No], sum(B.EXTDCOST) as [IV Total]
    from GL00105 A
    INNER JOIN SEE30303 B on A.ACTINDX = B.IVIVINDX
    group by A.ACTNUMST
) T1
INNER JOIN
(
    select A.ACTNUMST as [Account No], SUM(C.PERDBLNC) as [GL Total]
    from GL00105 A
    LEFT JOIN GL10110 C on A.ACTINDX = C.ACTINDX
    group by A.ACTNUMST
) T2
ON T1.[Account No] = T2.[Account No]

希望对你有帮助

【讨论】：

非常感谢。解决了我的问题。我遵循相同的逻辑，但尝试将其简化一点并得到我想要的结果。选择 T3.ACTNUMST 作为 [Account No], T1.[IV Total], T2.[GL Total] from (select IVIVINDX as [Index], SUM(EXTDCOST) as [IV Total] from SEE30303 group by IVIVINDX)T1 INNER JOIN （选择 ACTINDX 作为 [Index]，sum(PERDBLNC) 作为 [GL Total] from GL10110 group by ACTINDX）T2 on T1.[Index] = T2.[Index] INNER JOIN GL00105 T3 on T1.[Index] = T3.[ ACTINDX] 按 T3.ACTNUMST 排序
不确定如何在评论中输入代码部分？这个论坛的新手... 5 个空格似乎不起作用。
很高兴知道这一点。如果答案对您有用，请投票和/或标记为已接受。 :)
您可以使用反引号，即 ` 在大多数键盘的 Esc 按钮下方的注释中添加代码块。所以你的代码看起来像这样This is a code。在这个问题中不需要这样做，但您可以在以后的帖子中这样做。检查此链接meta.stackexchange.com/questions/74784/…
'select T3.ACTNUMST as [Account No], T1.[IV Total], T2.[GL Total] from (select IVIVINDX as [Index], SUM(EXTDCOST) as [IV Total]来自 IVIVINDX 的 SEE30303 组）T1 INNER JOIN（选择 ACTINDX 作为 [Index]，sum(PERDBLNC) 作为 [GL Total] from GL10110 group by ACTINDX）T2 on T1.[Index] = T2.[Index] INNER JOIN GL00105 T3 on T1.[Index] = T3.[ACTINDX] order by T3.ACTNUMST'

【解决方案2】：

这也可以通过多个 cte 来完成

 with IVTotal as
 (
     select A.ACTNUMST, sum(B.EXTDCOST) as [IV Total]
     from   GL00105 A
     INNER  JOIN SEE30303 B 
     on     A.ACTINDX = B.IVIVINDX
     group  by A.ACTNUMST
 ),  
 GLTotal as 
 (
     select A.ACTNUMST as [Account No], SUM(C.PERDBLNC) as [GL Total]
     from   GL00105 A
     LEFT   JOIN GL10110 C 
     on     A.ACTINDX = C.ACTINDX
     group  by A.ACTNUMST
 )
 select a.*,b.[GL Total] 
 from   IVTotal a ,GLTotal b
 where  a.ACTNUMST =  b.aCTNUMST

【讨论】：