django – 从结果中排除整个QuerySet

5b51 2022/1/14 8:22:52 python 字数 2175 阅读 532 来源 www.jb51.cc/python

我有以下型号： class LibraryEntry(models.Model): player = models.ForeignKey(Player) player_lib_song_id = models.IntegerField() title = models.CharField(max_length=200) artist = models.CharField(m

概述

class LibraryEntry(models.Model):
   player = models.ForeignKey(Player)
   player_lib_song_id = models.IntegerField()
   title = models.CharField(max_length=200)
   artist = models.CharField(max_length=200)
   album = models.CharField(max_length=200)
   track = models.IntegerField()
   genre = models.CharField(max_length=50)
   duration = models.IntegerField()
   is_deleted = models.BooleanField(default=False)

   class Meta:
     unique_together = ("player","player_lib_song_id")

   def __unicode__(self):
     return "Library Entry " + str(self.player_lib_song_id) + ": " + self.title

 class BannedSong(models.Model):
   lib_entry = models.ForeignKey(LibraryEntry)

   def __unicode__(self):
     return "Banned Library Entry " + str(self.lib_entry.title)

我想做这样的查询：

banned_songs = BannedSong.objects.filter(lib_entry__player=activePlayer)
 available_songs = LibraryEntry.objects.filter(player=activePlayer).exclude(banned_songs)

基本上如果一首歌被禁止,我想把它从我的可用歌曲中排除.有没有办法在Django中这样做？

banned_song_ids = (BannedSong.objects.filter(lib_entry__player=activePlayer)
                                            .values_list('lib_entry',flat=True))

available_songs = (LibraryEntry.objects.filter(player=activePlayer)
                                            .exclude('id__in' = banned_song_ids))

替代方案是：

available_songs = (LibraryEntry.objects.filter(player=activePlayer)
                                          .filter(bannedsong__isnull = True))

总结

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django – 从结果中排除整个QuerySet

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