6

我有一个简单的音乐模式：Artist、Release、Track 和 Song。前 3 个都是逻辑结构，而第四个（Song）是（Artist、Release、Track）作为 mp3、wav、ogg 等的特定实例。

我无法在数据库中生成有序的歌曲列表。问题是两者Track都有Release一个Artist. 虽然Song.Track.Artist始终是表演者姓名，Song.Track.Release.Artist但可能是表演者姓名或“各种艺术家”用于编译。我希望能够按其中一个进行排序，但我无法找出正确的方法来完成这项工作。

这是我的架构：

class Artist(models.Model):
    name = models.CharField(max_length=512)

class Release(models.Model):
    name = models.CharField(max_length=512)
    artist = models.ForeignKey(Artist)

class Track(models.Model):
    name = models.CharField(max_length=512)
    track_number = models.IntegerField('Position of the track on its release')
    length = models.IntegerField('Length of the song in seconds')
    artist = models.ForeignKey(Artist)
    release = models.ForeignKey(Release)

class Song(models.Model):
    bitrate = models.IntegerField('Bitrate of the song in kbps')
    location = models.CharField('Permanent storage location of the file', max_length=1024)
    owner = models.ForeignKey(User)
    track = models.ForeignKey(Track)

我的查询应该相当简单；过滤特定用户拥有的所有歌曲，然后按Song.Track.Artist.name或对它们进行排序Song.Track.Release.Artist.name。这是我在视图中的代码，按以下方式排序Song.Track.Artist.name：

songs = Song.objects.filter(owner=request.user).select_related('track__artist', 'track__release', 'track__release__artist').order_by('player_artist.name')

除非我order_by使用tblname.colname. 我查看了底层查询对象的as_sql方法，它表明当进行内连接以获取Song.Track.Release.Artist临时名称时，表T6使用了临时名称，Artist因为已经在同一个表上完成了内连接以获取Song.Track.Artist：

>>> songs = Song.objects.filter(owner=request.user).select_related('track__artist', 'track__release', 'track__release__artist').order_by('T6.name')
>>> print songs.query.as_sql()
('SELECT "player_song"."id", "player_song"."bitrate", "player_song"."location", 
  "player_song"."owner_id", "player_song"."track_id", "player_track"."id", 
  "player_track"."name", "player_track"."track_number", "player_track"."length", 
  "player_track"."artist_id", "player_track"."release_id", "player_artist"."id", 
  "player_artist"."name", "player_release"."id", "player_release"."name", 
  "player_release"."artist_id", T6."id", T6."name" FROM "player_song" INNER JOIN 
  "player_track" ON ("player_song"."track_id" = "player_track"."id") INNER JOIN 
  "player_artist" ON ("player_track"."artist_id" = "player_artist"."id") INNER JOIN 
  "player_release" ON ("player_track"."release_id" = "player_release"."id") INNER JOIN 
  "player_artist" T6 ON ("player_release"."artist_id" = T6."id") WHERE 
  "player_song"."owner_id" = %s  ORDER BY T6.name ASC', (1,))

当我将其作为表名放入时，order_by它确实有效（参见上面的示例输出），但这似乎完全不可移植。当然有更好的方法来做到这一点！我错过了什么？

4

1 回答 1

26

恐怕我真的无法理解你的问题是什么。

一些更正：select_related与排序无关（它根本不改变查询集，只是跟随连接来获取相关对象并缓存它们）；并按相关模型中的字段排序，您使用双下划线表示法，而不是点。例如：

Song.objects.filter(owner=request.user).order_by('track__artist__name')

但是在您的示例中，您使用了“player_artist”，这似乎不是模型中任何地方的字段。而且我不明白您提到的可移植性。

于 2009-10-23T10:07:03.490 回答