排除从结果

Question

整个查询集我有以下型号：

class LibraryEntry(models.Model): 
    player = models.ForeignKey(Player) 
    player_lib_song_id = models.IntegerField() 
    title = models.CharField(max_length=200) 
    artist = models.CharField(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) 

我愿意做这样的查询：

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

基本上如果一首歌曲被禁止，我想排除它来自我的可用歌曲集。有没有办法在Django中做到这一点？

Answer 1

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)) 

另一种方法是：

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