Question

我遇到了这个问题：

我有两张桌子：样本和歌曲我想选择

pathName   from samples
id_user    from samples
id         from songs
name       from songs

每个样本都属于一首歌：我想将每个共享相同歌曲ID的样本分组，以便选择最后一个日期。所以我在这两个表之间进行连接：

$query = 'SELECT 
            samples.pathName path_name,
            samples.id_user id_user,
            songs.id id_song,
            songs.name song_name
        FROM (SELECT *, MAX(date) AS maxDate FROM samples GROUP BY id_song ORDER BY maxDate) samples
        INNER JOIN songs songs
        ON songs.id = samples.id_chanson
        WHERE songs.finished = false';

它不应该那么重，我希望我很清楚......：）

Answer 1

尝试：

SELECT sa.pathName,
       sa.id_user,
       so.id id_song,
       so.name song_name
FROM songs so
JOIN (SELECT id_song, MAX(`date`) AS maxDate FROM samples GROUP BY id_song) mx
  ON so.id = mx.id_song
JOIN samples sa 
  on mx.id_song = sa.id_song and mx.maxDate = sa.`date`
WHERE so.finished = 'false'
ORDER BY mx.maxDate

Answer 2

我会用这样的东西：

SELECT
  samples.pathName path_name,
  samples.id_user id_user,
  songs.id id_song,
  songs.name song_name
FROM
  samples INNER JOIN songs ON songs.id = samples.id_song
WHERE
  songs.finished = false
  AND (samples.id_song, samples.`date`) IN (SELECT   id_song, MAX(`date`)
                                            FROM     samples
                                            GROUP BY id_song)

我如何在MySQL的子查询中按MAX（日期）进行ORDER？

2 个答案: