Mysql 用于5个表的SQL联接

我想加入5个表，我的查询是：

 SELECT maingenre.MainGenreId,maingenre.Name AS mname,'saewill.mp3' AS mp3,releaseprod.ArtistId,r me
FROM maingasdsadenre 
LEFT JOIN releasaddsaseprod ON maidssadngenre.MainGenreId = dd.a
LEFT JOIN artist ON d.ArtistId = d.ArtistId
l 

它没有给我一个准确的结果。 我有一个错误：

1066-非唯一表/别名：“releaseprod” 试试这个：-

SELECT maingenere.MainGenreId,maingenere.Name as mname,'saewill.mp3' as mp3,
       releaseprod.ArtistId,releaseprod.MainGenreId,releaseprod.LabelId,artist.ArtistId,
       artist.Name as aname,song.ArtistId,song.Name as sname,song.Version,label.LabelId,
       label.Name as lname
FROM maingenere 
left join releaseprod ON maingenere.MainGenreId = releaseprod.MainGenreId 
left join artist ON artist.ArtistId = releaseprod.ArtistId
left join song ON song.ArtistId = releaseprod.ArtistId
left join label ON label.LabelId = releaseprod.LabelId;

如果您的加入始终处于发布状态，请尝试此

或 编辑您的问题并告诉我编辑我的回答 或者你是这个意思

SELECT maingenere.MainGenreId,
  maingenere.Name AS mname,
  'saewill.mp3'   AS mp3,
  releaseprod.ArtistId,
  releaseprod.MainGenreId,
  releaseprod.LabelId,
  artist.ArtistId,
  artist.Name AS aname,
  song.ArtistId,
  song.Name AS sname,
  song.Version,
  label.LabelId,
  label.Name AS lname
FROM maingenere 
LEFT JOIN releaseprod
ON (maingenere.MainGenreId = releaseprod.MainGenreId)
LEFT JOIN artist
ON (artist.ArtistId = releaseprod.ArtistId)
LEFT JOIN song
ON (song.ArtistId = releaseprod.ArtistId)
LEFT JOIN label
ON( label.LabelId = releaseprod.LabelId)

SELECT maingenere.MainGenreId,
  maingenere.Name AS mname,
  'saewill.mp3'   AS mp3,
  releaseprod.ArtistId,
  releaseprod.MainGenreId,
  releaseprod.LabelId,
  artist.ArtistId,
  artist.Name AS aname,
  song.ArtistId,
  song.Name AS sname,
  song.Version,
  label.LabelId,
  label.Name AS lname
FROM maingenere 
LEFT JOIN releaseprod
ON (maingenere.MainGenreId = releaseprod.MainGenreId)
LEFT JOIN artist
ON (artist.ArtistId = releaseprod.ArtistId)
LEFT JOIN song
ON (song.ArtistId = releaseprod.ArtistId)
LEFT JOIN label
ON( label.LabelId = releaseprod.LabelId)