我有一个表(在SQL Server 2008中),它有垂直销售交易数据,我想将垂直输出转换为水平输出,我试图使用PIVOT,但有些人怎么不知道如何在PIVOT上应用group by ,因为我想要基于AccountHeadID,TransType和IsPast Column的Sum。
样本表
CREATE TABLE Trans
([TransID] int,
[CustID] int,
[AccountHeadID] int,
[TransType] varchar(100),
[IsPast] bit,
[Amount] int)
;
INSERT INTO Trans
([TransID],CustID, [AccountHeadID], [TransType], [IsPast],[Amount])
VALUES
(1, 1, 1, 'Sales',1,1000),
(2, 1, 1, 'Sales',0,500),
(3, 1, 1, 'Sales',0,400),
(4, 1, 1, 'Return',0,300),
(5, 1, 1, 'Return',0,200),
(6, 1, 1, 'Return',0,100),
(7, 1, 1, 'Return',1,150),
(8, 1, 2, 'Sales',1,2000)
当前查询
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT ',' + QUOTENAME(AccountHeadID)
from Trans
group by AccountHeadID
order by AccountHeadID
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT CustID,' + @cols + '
from
(
SELECT
a.CustID,
a.AccountHeadID,
a.TransType,
a.Amount,
a.isPast
FROM Trans a
) x
pivot
(
sum(Amount)
for AccountHeadID in (' + @cols + ')
) p '
execute sp_executesql @query;
预期输出
CustID | Account-HeadID-TransType-IsPast[1-Sales-Past] | Account-HeadID-TransType-IsCurrent[1-Sales-Current] | Account-HeadID-TransType-IsCurrent[1-Return-Past] | Account-HeadID-TransType-IsCurrent[1-Return-Current] | Account-HeadID-TransType-IsCurrent[2-Sales-Past]| ...
1 |1000 | 900 (500 + 400) |150 | 600[300+200+100] |2000
欢迎任何建议或意见!
由于 苏雷什
答案 0 :(得分:1)
试试这个:
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT ',' + QUOTENAME(CONVERT(NVARCHAR(10),AccountHeadID) + N'-' + TransType + N'-' + CONVERT(NVARCHAR(10),IsPast))
from Trans
group by AccountHeadID, TransType, IsPast
order by AccountHeadID, TransType, IsPast
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
--select @cols
set @query = 'SELECT CustID,' + @cols + '
from
(
SELECT
a.CustID,
CONVERT(NVARCHAR(10),AccountHeadID) + N''-'' + TransType + N''-'' + CONVERT(NVARCHAR(10),IsPast) Acct,
a.Amount
FROM Trans a
) x
pivot
(
sum(Amount)
for Acct in (' + @cols + ')
) p '
execute sp_executesql @query;