我如何得到:
id Name Value
1 A 4
1 B 8
2 C 9
to
id Column
1 A:4, B:8
2 C:9
我如何得到:
id Name Value
1 A 4
1 B 8
2 C 9
to
id Column
1 A:4, B:8
2 C:9
当前回答
使用Sql Server 2005及以上版本的另一种选择
---- test data
declare @t table (OUTPUTID int, SCHME varchar(10), DESCR varchar(10))
insert @t select 1125439 ,'CKT','Approved'
insert @t select 1125439 ,'RENO','Approved'
insert @t select 1134691 ,'CKT','Approved'
insert @t select 1134691 ,'RENO','Approved'
insert @t select 1134691 ,'pn','Approved'
---- actual query
;with cte(outputid,combined,rn)
as
(
select outputid, SCHME + ' ('+DESCR+')', rn=ROW_NUMBER() over (PARTITION by outputid order by schme, descr)
from @t
)
,cte2(outputid,finalstatus,rn)
as
(
select OUTPUTID, convert(varchar(max),combined), 1 from cte where rn=1
union all
select cte2.outputid, convert(varchar(max),cte2.finalstatus+', '+cte.combined), cte2.rn+1
from cte2
inner join cte on cte.OUTPUTID = cte2.outputid and cte.rn=cte2.rn+1
)
select outputid, MAX(finalstatus) from cte2 group by outputid
其他回答
如果group by只包含一个项目,您可以通过以下方式显著提高性能:
SELECT
[ID],
CASE WHEN MAX( [Name]) = MIN( [Name]) THEN
MAX( [Name]) NameValues
ELSE
STUFF((
SELECT ', ' + [Name] + ':' + CAST([Value] AS VARCHAR(MAX))
FROM #YourTable
WHERE (ID = Results.ID)
FOR XML PATH(''),TYPE).value('(./text())[1]','VARCHAR(MAX)')
,1,2,'') AS NameValues
END
FROM #YourTable Results
GROUP BY ID
让我们变得非常简单:
SELECT stuff(
(
select ', ' + x from (SELECT 'xxx' x union select 'yyyy') tb
FOR XML PATH('')
)
, 1, 2, '')
替换这一行:
select ', ' + x from (SELECT 'xxx' x union select 'yyyy') tb
你的疑问。
不需要游标,WHILE循环或用户定义函数。
只需要创造性地使用FOR XML和PATH。
[注意:此解决方案仅适用于SQL 2005及更高版本。原来的问题没有指定使用的版本。
CREATE TABLE #YourTable ([ID] INT, [Name] CHAR(1), [Value] INT)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (1,'A',4)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (1,'B',8)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (2,'C',9)
SELECT
[ID],
STUFF((
SELECT ', ' + [Name] + ':' + CAST([Value] AS VARCHAR(MAX))
FROM #YourTable
WHERE (ID = Results.ID)
FOR XML PATH(''),TYPE).value('(./text())[1]','VARCHAR(MAX)')
,1,2,'') AS NameValues
FROM #YourTable Results
GROUP BY ID
DROP TABLE #YourTable
一个例子是
在Oracle中可以使用LISTAGG聚合函数。
原始记录
name type
------------
name1 type1
name2 type2
name2 type3
Sql
SELECT name, LISTAGG(type, '; ') WITHIN GROUP(ORDER BY name)
FROM table
GROUP BY name
导致
name type
------------
name1 type1
name2 type2; type3
使用Replace函数和FOR JSON PATH
SELECT T3.DEPT, REPLACE(REPLACE(T3.ENAME,'{"ENAME":"',''),'"}','') AS ENAME_LIST
FROM (
SELECT DEPT, (SELECT ENAME AS [ENAME]
FROM EMPLOYEE T2
WHERE T2.DEPT=T1.DEPT
FOR JSON PATH,WITHOUT_ARRAY_WRAPPER) ENAME
FROM EMPLOYEE T1
GROUP BY DEPT) T3
有关示例数据和更多方法,请点击这里