Question

我有两张桌子！

歌曲：

|---------------------|------------------|------------------|
|      Title          |         ID       |      Version     |
|---------------------|------------------|------------------|
|   songnumber1       |         34       |    original      |
|---------------------|------------------|------------------|
|   songnumber2       |         35       |      remix       |
|---------------------|------------------|------------------|
|   songnumber3       |         36       |      remix       |
|---------------------|------------------|------------------|

我有其他表格可以在歌曲之间进行参考参考：

|---------------------|------------------|------------------|
|         ID          |      remix       |      original    |
|---------------------|------------------|------------------|
|         1           |        35        |        34        |
|---------------------|------------------|------------------|
|         2           |        36        |        34        |
|---------------------|------------------|------------------|

我想要的是显示当前歌曲下面的相关歌曲让我们说$ id是当前歌曲的ID，即35 我想显示所有其他歌曲混音和原版我尝试了这个，但如果$ id是remix：

，它只会得到原始版本

SELECT * FROM reference
         INNER JOIN songs ON reference.remix = songs.id OR 
reference.original = songs.id
         WHERE reference.original = '$id' OR reference.remix = '$id'

Answer 1

我认为这就是你想要的：

SELECT * FROM reference r
INNER JOIN songs a ON r.remix = a.ID  
INNER JOIN songs b ON r.original = b.ID
WHERE r.remix = 35

如果您发布预期结果会有所帮助。也许这就是你想要的：

SELECT * FROM reference r
INNER JOIN Songs a ON r.remix = a.ID  
WHERE r.remix = 35
UNION  
SELECT * FROM reference r
INNER JOIN Songs b ON r.original = b.ID
WHERE r.remix = 35

Answer 2

试试这个：

SELECT 
  C.currently_playing, 
  GROUP_CONCAT(DISTINCT IF(C.currently_playing=C.id,C.title,null)) current_song_title, 
  GROUP_CONCAT(DISTINCT IF(C.version='original',C.id,null)) orignial_song_id,
  GROUP_CONCAT(DISTINCT IF(C.version='original',C.title,null)) orignial_song_title, 
  GROUP_CONCAT(DISTINCT IF(C.version='remix',C.id,null)) related_remix_ids, 
  GROUP_CONCAT(DISTINCT IF(C.version='remix',C.title,null)) related_remix_titles 
FROM
  (SELECT 35 AS currently_playing, A.*
   FROM Songs   A 
   RIGHT JOIN (SELECT * FROM `references` WHERE remix=35 or original=35) B
   ON A.id IN (B.original,B.remix)) C
GROUP BY C.currently_playing;

查看有效的Demo on SQL Fiddle。

mysql获取所有其他行如何从同一个表中具有相同的id

2 个答案: