Question

我有三个简单的表格如下所示：

+----+-------------+  +----+-----------+  +----+---------------+
| artists          |  | albums         |  | songs              |
+----+-------------+  +----+-----------+  +----+---------------+
| id | name        |  | id | name      |  | id | name          |
+----+-------------+  +----+-----------+  +----+---------------+
| 0  | The Beatles |  | 0  | Help!     |  | 0  | Help!         |
| 1  | Metallica   |  | 1  | Metallica |  | 1  | Enter Sandman |
+----+-------------+  +----+-----------+  +----+---------------+

我希望查询结果如下：

+--------+---------------+
| type   | name          |
+--------+---------------+
| artist | The Beatles   |
| artist | Metallica     |
| album  | Help!         |
| album  | Metallica     |
| song   | Help!         |
| song   | Enter Sandman |
+--------+---------------+

我正在尝试使用LIKE查询来搜索艺术家，专辑和歌曲。

现在我使用下面的SQL从三列中获取结果，然后我使用Ruby将它们组合在一个对象中。我认为使用纯SQL而不是解释语言返回结果会更有效。

SELECT name FROM artists WHERE name LIKE 'something%'
SELECT name FROM albums WHERE name LIKE 'something%'
SELECT name FROM songs WHERE name LIKE 'something%'

我对SQL并不是很好，并且几天来一直在努力解决这个问题，并且想知道这里是否有人能指出我正确的方向。

Answer 1

您可以使用UNION将类似的结果合并在一起。试试这个：

SELECT 'artist' as Type, name FROM artists WHERE name LIKE 'something%'
UNION SELECT 'album' as Type, name FROM albums WHERE name LIKE 'something%'
UNION SELECT 'song' as Type, name FROM songs WHERE name LIKE 'something%'

Answer 2

您应该使用union在单个结果集中合并结果，并根据表格对您想要的类型进行硬编码。

SELECT 'artist' as [type], name FROM artists WHERE name LIKE 'something%'
UNION
SELECT 'album' as [type], name FROM albums WHERE name LIKE 'something%'
UNION
SELECT 'song' as [type], name FROM songs WHERE name LIKE 'something%'

Answer 3

使用带有常量字符串的union作为结果的列，例如

      select 'artist' as type, name as name from artists
union select 'album' as type, name as name from albums
union select 'song' as type, name as name from songs

Answer 4

您可以像其他人所建议的那样使用UNION。但是，UNION的语义强制数据库消除通常需要排序操作的重复行。由于您知道每个查询都必须返回非重复行，因此使用UNION ALL会更有效。像

这样的东西

SELECT *
  FROM (SELECT 'artists' as type, name 
          FROM artists 
        UNION ALL
        SELECT 'albums' as type, name 
          FROM albums
        UNION ALL
        SELECT 'songs' as type, name 
          FROM songs)
 WHERE name LIKE 'something%'

无论您使用什么数据库引擎，都应该能够将NAME上的谓词推送到一起进行UNION ALL的三个分支中的每个分支。但是在查询计划中确实值得确认。

Answer 5

select 'artists' as type, name from artists where name like '%something%'
union 
select 'albums' , name from albums where ...
union 
select 'songs', name from songs where ...

组合多列

5 个答案: