Question

我有2个Django模特 - DJ和歌曲。

class DJ(models.Model):    
    name = models.CharField(max_length=50, unique=True)
    rank = models.IntegerField()

class Song(models.Model):
    song_id = models.IntegerField()
    name = models.CharField(max_length=150)
    artist = models.ForeignKey(DJ, default=None, blank=True, null=True)
    release = models.CharField(max_length=500, default=None, blank=True, null=True)
    release_date = models.DateField()
    duplicate = models.BooleanField(default=False)

我试图获取特定DJ的所有同名歌曲。

这就是我这样做的方式

for dj in DJ.objects.all():
    song_group_list = []
    dj_song_list = Song.objects.filter(artist=dj)
    for song in dj_song_list:
        song_group = dj_song_list.filter(name=song.name).order_by('song_id')
        if len(song_group) > 1:
            if song_group not in song_group_list:
                song_group_list.append(song_group)
    for group in song_group_list:
        print group

这将输出具有相同查询集结果的重复集。

[<Song: Ping Pong>, <Song: Ping Pong>, <Song: Ping Pong>]
[<Song: Everybody Is In The Place>, <Song: Everybody Is In The Place>, <Song: Everybody Is In The Place>]
[<Song: Jumper>, <Song: Jumper>, <Song: Jumper>]
[<Song: Man With The Red Face>, <Song: Man With The Red Face>, <Song: Man With The Red Face>, <Song: Man With The Red Face>, <Song: Man With The Red Face>, <Song: Man With the Red Face>]
[<Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>]
[<Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>, <Song: Wake Up>]

如何针对上述问题获取不同的查询集？

Answer 1

我认为这个问题是因为append方法，因为，首先你得到所有的os歌曲列表，然后将它附加到song_list，所以你必须用有序列表替换当前列表：

所以请将您的代码更改为：

for dj in DJ.objects.all():
    song_group_list = []
    dj_song_list = Song.objects.filter(artist=dj)
    for song in dj_song_list:
        song_group = dj_song_list.filter(name=song.name).order_by('song_id')
        if len(song_group) > 1 and (song_group not in song_group_list):
                song_group_list=song_group
    for group in song_group_list:
        print group

如何从Django中的一组查询集中获取不同的查询集？

1 个答案: