我对这个查询有问题:
我有两张桌子:我想选择的样本和歌曲
pathName from samples
id_user from samples
id from songs
name from songs
每个样本都属于一首歌曲:我想将共享相同歌曲 ID 的每个样本分组,以便选择最后一个日期。所以我像这样在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...... :)