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