在JOIN

Question

Table_One：

+---------+----------+
|  USER_ID|   name   |
+---------+----------+
|    2    |   Rick   |
+---------+----------+

Table_Two

+---------+----------+----------+
|  t2_ID  |  sport   |  USER_ID |
+---------+----------+----------+
|    32   | football |     2    |
+---------+----------+----------+
|    33   | baseball |     2    |
+---------+----------+----------+

我希望得到的结果如下：

+---------+----------+--------------------+
|    2    |  Rick    | Football, Baseball |
+---------+----------+--------------------+

我不明白如何将这样一个INNER JOIN的多行合并为一列。

SELECT Table_One.* FROM Table_One LEFT JOIN Table_Two ON Table_Two.USER_ID = Table_One.USER_ID

我将结果作为单独的行：

+---------+----------+----------+
|    2    |  Rick    | football |
+---------+----------+----------+
|    2    |  Rick    | baseball |
+---------+----------+----------+

谢谢所有阅读此书的人。

Answer 1

在进行任何评论/编辑之前，此问题基于标签回答。

您可以使用FOR XML PATH()子句：

SELECT to.*,
       STUFF( (SELECT DISTINCT ', ' +tt.sport   
               FROM Table_Two tt
               WHERE tt.USER_ID = to.USER_ID
               FOR XML PATH('')
              ), 1, 1, ''
           ) AS Sports
FROM Table_One to;