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
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
其他回答
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
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 '%,%'
这只是Kevin Fairchild的文章的补充(顺便说一句,非常聪明)。我本来会把它作为一个评论,但我还没有足够的分数:)
我将这个想法用于我正在工作的视图,然而我正在连接的项目包含空间。因此,我稍微修改了代码,不再使用空格作为分隔符。
再次感谢你酷炫的解决办法,凯文!
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],
REPLACE(REPLACE(REPLACE(
(SELECT [Name] + ':' + CAST([Value] AS VARCHAR(MAX)) as A
FROM #YourTable
WHERE ( ID = Results.ID )
FOR XML PATH (''))
, '</A><A>', ', ')
,'<A>','')
,'</A>','') AS NameValues
FROM #YourTable Results
GROUP BY ID
DROP TABLE #YourTable
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
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