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

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

其他回答

当我尝试将Kevin Fairchild的建议转换为使用包含空格和编码的特殊XML字符(&,<,>)的字符串时,遇到了几个问题。

我的代码的最终版本(它没有回答最初的问题,但可能对某些人有用)看起来像这样:

CREATE TABLE #YourTable ([ID] INT, [Name] VARCHAR(MAX), [Value] INT)

INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (1,'Oranges & Lemons',4)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (1,'1 < 2',8)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (2,'C',9)

SELECT  [ID],
  STUFF((
    SELECT ', ' + CAST([Name] AS VARCHAR(MAX))
    FROM #YourTable WHERE (ID = Results.ID) 
    FOR XML PATH(''),TYPE 
     /* Use .value to uncomment XML entities e.g. &gt; &lt; etc*/
    ).value('.','VARCHAR(MAX)') 
  ,1,2,'') as NameValues
FROM    #YourTable Results
GROUP BY ID

DROP TABLE #YourTable

它没有使用空格作为分隔符并将所有空格替换为逗号,而是在每个值前附加一个逗号和空格,然后使用STUFF删除前两个字符。

XML编码由TYPE指令自动处理。

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

从http://groupconcat.codeplex.com安装SQLCLR聚合

然后你可以像这样写代码来得到你想要的结果:

CREATE TABLE foo
(
 id INT,
 name CHAR(1),
 Value CHAR(1)
);

INSERT  INTO dbo.foo
    (id, name, Value)
VALUES  (1, 'A', '4'),
        (1, 'B', '8'),
        (2, 'C', '9');

SELECT  id,
    dbo.GROUP_CONCAT(name + ':' + Value) AS [Column]
FROM    dbo.foo
GROUP BY id;

我使用了这种方法,可能更容易掌握。获取一个根元素,然后连接到具有相同ID但不是“正式”名称的选项

  Declare @IdxList as Table(id int, choices varchar(max),AisName varchar(255))
  Insert into @IdxLIst(id,choices,AisName)
  Select IdxId,''''+Max(Title)+'''',Max(Title) From [dbo].[dta_Alias] 
 where IdxId is not null group by IdxId
  Update @IdxLIst
    set choices=choices +','''+Title+''''
    From @IdxLIst JOIN [dta_Alias] ON id=IdxId And Title <> AisName
    where IdxId is not null
    Select * from @IdxList where choices like '%,%'

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