简单问题:
有 2 个表:
流派[名称、歌曲ID]
歌曲 [id、标题、用户 ID、状态]
从歌曲 INNER JOIN 流派中选择 ID、名称、标题、用户 ID、状态 Song.id=genre.songID ORDER BY id ASC；
从什么查询中获取结果
+----+-------------+----------------------+--------+--------+
| id | genre.name | song.title | userID | status |
+----+-------------+----------------------+--------+--------+
| 1 | tech | Feel it all | 1 | 1 |
| 2 | tech | Tester | 1 | 1 |
| 3 | music | Sejujurnya | 1 | 1 |
| 4 | music | Not Done | 1 | 1 |
| 5 | life | Cinta | 1 | 1 |
| 6 | life | Feel it all | 1 | 1 |
| 7 | life | Not Done | 1 | 1 |
| 8 | truth | Tester | 1 | 1 |
| 9 | tree | Tester | 1 | 1 |
| 10 | climb | Tester | 1 | 1 |
+----+-------------+----------------------+--------+--------+
至
+----+-------------+---------------------------------+--------+--------+
| id | genre.name | song.title | userID | status |
+----+-------------+---------------------------------+--------+--------+
| 1 | tech | Feel it all,Tester | 1 | 1 |
| 2 | music | Sejujurnya, Not Done | 1 | 1 |
| 3 | life | Cinta, Feel it all, Note Done | 1 | 1 |
| 4 | truth | Tester | 1 | 1 |
| 5 | tree | Tester | 1 | 1 |
| 6 | climb | Tester | 1 | 1 |
+----+-------------+---------------------------------+--------+--------+
谢谢
最佳答案
使用GROUP_CONCAT与 GROUP BY
SELECT
id,
genre.name,
GROUP_CONCAT(title) as title,
userID,
status
FROM
songs
INNER JOIN
genre
ON
song.id=genre.songID
GROUP BY
genre.name
ORDER BY
id ASC
关于php - 将一个 SELECT 组合到另一个 SELECT 中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18408570/