(SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6)as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleOFALL
GROUP BY ID_OF, Col, BNT, ref)
(SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6)as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleALL
GROUP BY ID_OF, Col, BNT, ref)
我在SQL Server中执行了此查询并获得此结果
Id_OF Col BNT size1 size2 size3 size4 size5 size6 size7 size8 size9 size10 Total ref
---------
37623 738 A 60 60 60 30 30 0 0 0 0 0 240 131380
而且这个：
Id_OF Col BNT size1 size2 size3 size4 size5 size6 size7 size8 size9 size10 Total ref
---------
37623 738 A 60 60 60 30 28 0 0 0 0 0 238 131380
如何在查询中减去这两个结果！我应该得到这个结果
Id_OF Col BNT size1 size2 size3 size4 size5 size6 size7 size8 size9 size10 Total ref
---------
37623 738 A 0 0 0 0 2 0 0 0 0 0 2 131380
谢谢堆
答案 0 :(得分：1)
存在许多方法，具体取决于您的实际逻辑，但对于您提出的问题，我只想使用公共表表达式，子查询或临时表将它们连接在一起，然后从相应的大小列中减去size列
这样的事情：
;WITH T1 AS
(
SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6)as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleOFALL
GROUP BY ID_OF, Col, BNT, ref) ,
T2 AS
(SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6)as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleALL
GROUP BY ID_OF, Col, BNT, ref)
SELECT
T1.size1 - t2.size1
.....
.....
.....
FROM T1
INNER JOIN T2 ON T1.ID_OF = T2.ID_OF
如果外部联接在两个查询之间不匹配，则使用外部联接 （例子只是为了说明）
但正如所说的那样 - 存在这样做的多种方式。
你也可以做一个查询 - 我怀疑 - 并在SUM和GROUP BY之前减去它。
答案 1 :(得分：1)
这看起来很熟悉...... 但是这个怎么样？
SELECT a.ID_OF, a.Col, a.BNT,
SUM(a.size1) - SUM(b.size1) As size1,
SUM(a.size2) - SUM(b.size2) As size2,
SUM(a.size3) - SUM(b.size3) As size3,
SUM(a.size4) - SUM(b.size4) As size4,
SUM(a.size5) - SUM(b.size5) As size5,
SUM(a.size6) - SUM(b.size6) As size6,
SUM(a.size7) - SUM(b.size7) As size7,
SUM(a.size8) - SUM(b.size8) As size8,
SUM(a.size9) - SUM(b.size9) As size9,
SUM(a.size10) - SUM(b.size10) As size10,
SUM(a.total) - SUM(b.total) As total,
a.ref
FROM tblTailleOFALL a
JOIN tblTailleALL b
ON a.ID_OF = b.ID_OF
AND a.Col = b.Col
AND a.BNT = b.BNT
GROUP BY a.ID_OF, a.Col, a.BNT, a.ref
答案 2 :(得分：1)
您可以加入表并减去值：
SELECT coalesce(a.ID_OF ,b.id) as id
, coalesce(a.Col, b.Col) as Col
, coalesce(a.BNT, b.BNT) as BNT
, SUM(a.size1 -b.size1 ) As size1
, SUM(a.size2 -b.size2 ) As size2
, SUM(a.size3 -b.size3 ) as size3
, SUM(a.size4 -b.size4 ) as size4
, SUM(a.size5 -b.size5 ) as size5
, SUM(a.size6 -b.size6 ) as size6
, SUM(a.size7 -b.size7 ) as size7
, SUM(a.size8 -b.size8 ) as size8
, SUM(a.size9 -b.size9 ) as size9
, SUM(a.size10-b.size10) as size10
, SUM(a.Total -b.Total ) as Total
, coalesce (a.ref, b.ref) as ref
FROM tblTailleOFALL a
full outer join
tblTailleALL b
on ( a.id = b.id
and a.Col = b.Col
and a.BNT = b.BNT
and a.ref = b.ref)
group by
coalesce(a.ID_OF ,b.id)
, coalesce(a.Col, b.Col)
, coalesce(a.BNT, b.BNT)
, coalesce (a.ref, b.ref)
答案 3 :(得分：0)
只需通过ID_OF加入这两个查询并减去外部查询中的每一列：
SELECT a.ID_OF,
a.Col,
a.BNT,
a.size1 - b.size1 AS size1,
a.size2 - b.size2 AS size2,
a.size3 - b.size3 AS size3,
a.size4 - b.size4 AS size4,
a.size5 - b.size5 AS size5,
a.size6 - b.size6 AS size6,
a.size7 - b.size7 AS size7,
a.size8 - b.size8 AS size8,
a.size9 - b.size9 AS size9,
a.size19 - b.size10 AS size10,
a.total - b.total AS total,
a.ref
FROM (
SELECT ID_OF,
Col,
BNT,
SUM(size1) AS size1,
SUM(size2) AS size2,
SUM(size3) AS size3,
SUM(size4) AS size4,
SUM(size5) AS size5,
SUM(size6) AS size6,
SUM(size7) AS size7,
SUM(size8) AS size8,
SUM(size9) AS size9,
SUM(size10) AS size10,
SUM(Total) AS Total,
ref
FROM tblTailleOFALL
GROUP BY ID_OF,
Col,
BNT,
ref
) a
INNER JOIN (
SELECT ID_OF,
Col,
BNT,
SUM(size1) AS size1,
SUM(size2) AS size2,
SUM(size3) AS size3,
SUM(size4) AS size4,
SUM(size5) AS size5,
SUM(size6) AS size6,
SUM(size7) AS size7,
SUM(size8) AS size8,
SUM(size9) AS size9,
SUM(size10) AS size10,
SUM(Total) AS Total,
ref
FROM tblTailleALL
GROUP BY ID_OF,
Col,
BNT,
ref
) b ON a.ID_OF = b.ID_OF
答案 4 :(得分：0)
SELECT a.ID_OF, a.Col, a.BNT,
a.size1 - b.size1 as size1,
a.size2 - b.size2 as size2,
a.size3 - b.size3 as size3,
a.size4 - b.size4 as size4,
a.size5 - b.size5 as size5,
a.size6 - b.size6 as size6,
a.size7 - b.size7 as size7,
a.size8 - b.size8 as size8,
a.size9 - b.size9 as size9,
a.size10 - b.size10 as size10,
a.Total - b.Total as Total,
a.ref
(SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6)as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleOFALL
GROUP BY ID_OF, Col, BNT, ref) AS a
inner join
(SELECT ID_OF, Col, BNT,
SUM(size1) As size1,
SUM(size2) As size2,
SUM(size3) as size3,
SUM(size4) as size4,
SUM(size5) as size5,
SUM(size6) as size6,
SUM(size7) as size7,
SUM(size8) as size8,
SUM(size9) as size9,
SUM(size10) as size10,
SUM(Total) as Total,ref
FROM tblTailleALL
GROUP BY ID_OF, Col, BNT, ref) AS b on a.ID_OF = b.ID_OF