而不是独立地连接多个表,使用单独的查询？

Question

我经常发现自己在桌子上进行了几次独立的连接.例如,假设我们有桌子collections,有独立的与两个一对N的关系photos和songs,其中N是从零到很多.

现在,假设我们希望得到一个集合,这两个公司(独立)相关联的照片和歌曲.

我通常会使用这样的东西:

SELECT
    collections.collectionid as collectionid,
    photos.name as photo_name,
    songs.name  as song_name

FROM collections
    LEFT JOIN photos ON collections.collectionid = photos.collectionid
    LEFT JOIN songs  ON collections.collectionid = songs.collectionid

WHERE collections.collectionid = 14

当然,将一个表左连接到另外两个表,如果第一个连接导致M行而第二个连接N成行,则给M * N出行.就数据库流量和性能而言,这似乎不是最理想的.

+--------------+------------+-----------+
| collectionid | photo_name | song_name |
+--------------+------------+-----------+
| 14           | 'x'        | 'a'       | \
| 14           | 'x'        | 'b'       |  - Each photo is returned 3 times,
| 14           | 'x'        | 'c'       | /  because 3 songs are returned.
| 14           | 'y'        | 'a'       | \
| 14           | 'y'        | 'b'       | 
| 14           | 'y'        | 'c'       | /
+--------------+------------+-----------+

或者,您可以执行两个选择:两个单独的查询,每个查询连接collections到不同的表,给M + N出行:

SELECT
    collections.collectionid as collectionid
    song.name as song_name
FROM collections
    LEFT JOIN songs on collections.collectionid = songs.collectionid
WHERE collections.collectionid = 14

和:

SELECT
    collections.collectionid as collectionid
    photos.name as photo_name
FROM collections
    LEFT JOIN photos on collections.collectionid = photos.collectionid
WHERE collections.collectionid = 14

赠送:

+--------------+------------+    +--------------+------------+
| collectionid | song_name  |    | collectionid | photo_name |
+--------------+------------+    +--------------+------------+
| 14           | 'a'        |    | 14           | 'x'        |
| 14           | 'b'        |    | 14           | 'y'        |
| 14           | 'c'        |    +--------------+------------+
+--------------+------------+

我的问题:处理这个问题的最佳方法是什么？

上述两者似乎都不是最佳的.那么,是否存在导致M + N行的另一种方式,但是可以在单个查询中完成？

Answer 1

您的第一个选项(两个独立的JOIN)似乎没有为您提供非常有用的结果集(因为这两个子表生成半笛卡尔积,您必须在应用程序代码中重复删除结果).

第二个选项(两个单独的查询)是可以的,除非您想将两个查询的结果视为单个集合用于表示目的(例如,通过日期字段将它们一起排序).

我认为,最好的解决方案是将两个查询合并为一个UNION ALL,生成一个只包含您实际需要的行的结果集:

SELECT
  collections.collectionid as collectionid,
  photos.name as photo_name,
  'photo' as document_type
FROM collections
  LEFT JOIN photos on collections.collectionid = photos.collectionid
WHERE collections.collectionid = 14
UNION ALL
SELECT
  collections.collectionid as collectionid,
  song.name as photo_name
  'song' as document_type
FROM collections
  LEFT JOIN songs on collections.collectionid = songs.collectionid
WHERE collections.collectionid = 14

这种结果集可以是ORDERed BY整个组合记录集中的任何字段,允许(例如)获取附加到集合的20个最新文档,而不管它们是什么类型.