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问题描述

我有两个名为 DistrictsSchools 的表.Districts 表包含一个名为 Schools 的列.

I have two tables named Districts and Schools. The Districts table contains a column named Schools.

我需要从对应的Schools表中填充Districts表的Schools列,以便中的每一行Districts 表有一个逗号分隔的列表,其中包含来自 Schools 表的学校名称值.

I need to populate the Schools column of the Districts table from the corresponding Schools table, so that each row in the Districts table has a comma separated list of values of school names from the Schools table.

我该怎么做?我应该使用 UPDATE 查询还是存储过程?

How can I do this? Should I use an UPDATE query or a stored procedure?

我只做到了:

SQL 小提琴

地区表

+------------+------+---------+
| DistrictId | Name | Schools |
+------------+------+---------+
|          1 | a    |         |
|          2 | b    |         |
|          3 | c    |         |
|          4 | d    |         |
+------------+------+---------+

学校表

+----------+------------+------------+
| SchoolId | SchoolName | DistrictId |
+----------+------------+------------+
|        1 | s1         |          1 |
|        2 | s2         |          1 |
|        3 | s3         |          2 |
|        4 | s4         |          2 |
|        5 | s5         |          4 |
+----------+------------+------------+

输出需要如何

+------------+------+---------+
| DistrictId | Name | Schools |
+------------+------+---------+
|          1 | a    | s1,s2   |
|          2 | b    | s3,s4   |
|          3 | c    |         |
|          4 | d    | s5      |
+------------+------+---------+

推荐答案

借助 FOR XML PATHSTUFF 来连接值,您可以使用您想要的结果轻松更新表 District.

UPDATE  a
SET     a.Schools = b.SchoolList
FROM    Districts a
        INNER JOIN
        (
            SELECT  DistrictId,
                    STUFF((SELECT ', ' + SchoolName
                            FROM Schools
                            WHERE DistrictId = a.DistrictId
                            FOR XML PATH (''))
                        , 1, 1, '')  AS SchoolList
            FROM    Districts AS a
            GROUP   BY DistrictId
        ) b ON A.DistrictId = b.DistrictId
WHERE   b.SchoolList IS NOT NULL

  • SQLFiddle 演示
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10-19 09:01