[英]SQL Query Second sum column squaring results
如果我分別運行2個查詢,我將得到如下結果。
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
第二查詢
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
我希望能夠將兩個結果結合在一起進行比較,但是我相信它會重復GL總和中每一行的總和,然后再次對其求和,得出的數字很大,例如-
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
什么時候應該
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
請告知如何使用求和功能以獲得正確的結果。
你想要這樣的東西嗎? 如果這不是預期的,請檢查並評論。
另外,請糾正所有語法錯誤。
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]
希望這可以幫助
這也可以通過多個CTE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
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.