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
另一个不带垃圾的例子:",TYPE).value('(./text())[1]','VARCHAR(MAX)')"
WITH t AS (
SELECT 1 n, 1 g, 1 v
UNION ALL
SELECT 2 n, 1 g, 2 v
UNION ALL
SELECT 3 n, 2 g, 3 v
)
SELECT g
, STUFF (
(
SELECT ', ' + CAST(v AS VARCHAR(MAX))
FROM t sub_t
WHERE sub_t.g = main_t.g
FOR XML PATH('')
)
, 1, 2, ''
) cg
FROM t main_t
GROUP BY g
输入-输出是
************************* -> *********************
* n * g * v * * g * cg *
* - * - * - * * - * - *
* 1 * 1 * 1 * * 1 * 1, 2 *
* 2 * 1 * 2 * * 2 * 3 *
* 3 * 2 * 3 * *********************
*************************
其他回答
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
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. > < 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。
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
不需要光标…while循环就足够了。
------------------------------
-- Setup
------------------------------
DECLARE @Source TABLE
(
id int,
Name varchar(30),
Value int
)
DECLARE @Target TABLE
(
id int,
Result varchar(max)
)
INSERT INTO @Source(id, Name, Value) SELECT 1, 'A', 4
INSERT INTO @Source(id, Name, Value) SELECT 1, 'B', 8
INSERT INTO @Source(id, Name, Value) SELECT 2, 'C', 9
------------------------------
-- Technique
------------------------------
INSERT INTO @Target (id)
SELECT id
FROM @Source
GROUP BY id
DECLARE @id int, @Result varchar(max)
SET @id = (SELECT MIN(id) FROM @Target)
WHILE @id is not null
BEGIN
SET @Result = null
SELECT @Result =
CASE
WHEN @Result is null
THEN ''
ELSE @Result + ', '
END + s.Name + ':' + convert(varchar(30),s.Value)
FROM @Source s
WHERE id = @id
UPDATE @Target
SET Result = @Result
WHERE id = @id
SET @id = (SELECT MIN(id) FROM @Target WHERE @id < id)
END
SELECT *
FROM @Target
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