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
其他回答
SQL Server 2005及其后续版本允许您创建自己的自定义聚合函数,包括像连接这样的功能—请参阅链接文章底部的示例。
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。
如果你启用了clr,你可以使用GitHub中的Group_Concat库
使用Stuff和for xml路径操作符将行连接到字符串:Group By两列——>
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 (1,'B',5)
INSERT INTO #YourTable ([ID],[Name],[Value]) VALUES (2,'C',9)
-- retrieve each unique id and name columns and concatonate the values into one column
SELECT
[ID],
STUFF((
SELECT ', ' + [Name] + ':' + CAST([Value] AS VARCHAR(MAX)) -- CONCATONATES EACH APPLICATION : VALUE SET
FROM #YourTable
WHERE (ID = Results.ID and Name = results.[name] )
FOR XML PATH(''),TYPE).value('(./text())[1]','VARCHAR(MAX)')
,1,2,'') AS NameValues
FROM #YourTable Results
GROUP BY ID
SELECT
[ID],[Name] , --these are acting as the group by clause
STUFF((
SELECT ', '+ CAST([Value] AS VARCHAR(MAX)) -- CONCATONATES THE VALUES FOR EACH ID NAME COMBINATION
FROM #YourTable
WHERE (ID = Results.ID and Name = results.[name] )
FOR XML PATH(''),TYPE).value('(./text())[1]','VARCHAR(MAX)')
,1,2,'') AS NameValues
FROM #YourTable Results
GROUP BY ID, name
DROP TABLE #YourTable
八年后……Microsoft SQL Server vNext数据库引擎最终增强了Transact-SQL,以直接支持分组字符串连接。社区技术预览版本1.0添加了STRING_AGG函数,CTP 1.1为STRING_AGG函数添加了WITHIN GROUP子句。
参考:https://msdn.microsoft.com/en-us/library/mt775028.aspx