结合多列

我有三个简单的表，如下图所示：结合多列

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

而且我想查询导致以下：

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

我试图用一个LIKE查询搜索艺术家，专辑和歌曲。

现在我使用下面的SQL从三列中获取结果，然后使用Ruby将它们放在一个对象中。我认为使用纯SQL而不是解释语言来返回结果会更有效率。

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

我没有与SQL如此之大，并已与此挣扎了几天，想知道如果在这里任何人都可以在正确的方向指向我。

来源

2011-06-20 RyanScottLewis

您可以像其他人所建议的那样使用UNION。然而，UNION的语义强制数据库消除通常需要排序操作的重复行。既然你知道每个查询都必须返回非重复的行，那么使用UNION ALL会更有效率。喜欢的东西

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%'

无论数据库引擎使用的是应该能够推动谓词NAME到各三个分支是正在UNION ALL'ed在一起。但是在查询计划中值得确认。

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2011-06-20 02:29:14

我正在考虑如何与其他答案完全一致。好样的！ – RyanScottLewis

您可以使用UNION将类似的结果结合在一起。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%'

来源

2011-06-20 01:32:00

我以前实际遇到过这个答案。这里的问题是我需要知道每个结果实际在哪个表中。尽管感谢您的快速回答！编辑：Bah，我错过了查询的“AS”部分。 = p – RyanScottLewis

您应该使用工会单个ResultSet合并的结果，同时也硬编码您可以根据表要的类型..

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%'

来源

2011-06-20 01:32:47

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 ...

来源

2011-06-20 01:33:42 Kal

固定字符串使用union作为结果的列，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

来源

2011-06-20 01:34:36

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