Question

我在实现连接多行的XML PATH方法时遇到了一些麻烦。因此，如下表所示，测试。

ID  Type  
 1  boy
 2  boy
 3  boy
 1  girl
 2  girl
 3  man
 3  woman

查询是：

SELECT DISTINCT a.ID,
    (
        SELECT  b.Type + ','
        FROM Test as b
        WHERE a.Type = b.Type
        for XML PATH ('')
    )
FROM Test as a

但不是返回：

ID  Type  
 1  boy,girl,man,
 2  boy,girl,
 3  boy,girl,woman

它改为返回：

ID  Type
 1  boy,boy,boy,
 1  girl,girl,
 2  boy,boy,boy,
 2  girl,girl,
 3  boy,boy,boy,
 3  man,
 3  woman,

发生了什么？

Answer 1

你加入了错误的领域。

尝试

SELECT DISTINCT a.ID,
    (
        SELECT  b.Type + ','
        FROM Test as b
        WHERE a.ID = b.ID
        for XML PATH ('')
    )
FROM Test as a

Answer 2

而不是使用DISTINCT更喜欢使用GROUP BY之类的东西......

SELECT  a.ID,

   STUFF((SELECT  ', ' +  [Type] [text()]
          FROM Test 
          WHERE ID = a.[ID]
          for XML PATH (''),TYPE).
          value('.','NVARCHAR(MAX)'),1,2,'') AS [Type]

FROM Test as a
GROUP BY a.ID

使用XML PATH连接多个字段

2 个答案: