如何实现Django中的存在？

Question

我在Django有两个模型，一个用于歌曲，一个用于专辑，一个专辑有很多歌曲。我正在尝试过滤歌曲有效的专辑。例如，至少有一首歌必须有一个音频文件，以便过滤器返回相册。我正在使用Postgres。

我试图通过Django QuerySet弄清楚如何做这个逻辑，但我不确定如何使用存在而不是存在。

以下是我想要开始工作的Django orm声明：

valid_songs = Song.objects.filter(
    album=OuterRef('pk'),
    audio_file__isnull=False).only("album")

Album.objects.annotate(
    valid_song=Exists(valid_songs)).filter(
valid_song=True).query

这是生成的查询：

SELECT "api_album"."id", 
       "api_album"."created_at", 
       "api_album"."updated_at", 
       "api_album"."title", 
       "api_album"."artwork_file_id", 
       "api_album"."user_id", 
       "api_album"."description", 
       "api_album"."tags", 
       "api_album"."genres", 
       EXISTS(SELECT U0."id", 
                     U0."album_id" 
              FROM   "api_song" U0 
              WHERE  ( U0."album_id" = ( "api_album"."id" ) 
                       AND U0."audio_file_id" IS NOT NULL )) AS "valid_song" 
FROM   "api_album" 
WHERE  EXISTS(SELECT U0."id", 
                     U0."album_id" 
              FROM   "api_song" U0 
              WHERE  ( U0."album_id" = ( "api_album"."id" ) 
                       AND U0."audio_file_id" IS NOT NULL )) = true 

这是由Django的QuerySet生成的上述查询的postgres查询计划：

Seq Scan on api_album  (cost=0.00..287.95 rows=60 width=641)
 Filter: (alternatives: SubPlan 3 or hashed SubPlan 4)
 SubPlan 3
   ->  Seq Scan on api_song u0_2  (cost=0.00..1.54 rows=1 width=0)
         Filter: ((audio_file_id IS NOT NULL) AND (album_id = api_album.id))
 SubPlan 4
   ->  Seq Scan on api_song u0_3  (cost=0.00..1.43 rows=10 width=4)
         Filter: (audio_file_id IS NOT NULL)
 SubPlan 1
   ->  Seq Scan on api_song u0  (cost=0.00..1.54 rows=1 width=0)
         Filter: ((audio_file_id IS NOT NULL) AND (album_id = api_album.id))
 SubPlan 2
   ->  Seq Scan on api_song u0_1  (cost=0.00..1.43 rows=10 width=4)
         Filter: (audio_file_id IS NOT NULL)
(14 rows)

但是，对此有更有效的查询

SELECT * 
FROM   "api_album" 
WHERE  EXISTS(SELECT U0."id", 
                     U0."album_id" 
              FROM   "api_song" U0 
              WHERE  ( U0."album_id" = ( "api_album"."id" ) 
                       AND U0."audio_file_id" IS NOT NULL )) 

Hash Semi Join  (cost=1.55..13.26 rows=10 width=640)
 Hash Cond: (api_album.id = u0.album_id)
 ->  Seq Scan on api_album  (cost=0.00..11.20 rows=120 width=640)
 ->  Hash  (cost=1.43..1.43 rows=10 width=4)
       ->  Seq Scan on api_song u0  (cost=0.00..1.43 rows=10 width=4)
             Filter: (audio_file_id IS NOT NULL)
(6 rows)

所以我的问题如下：

1. 在这种情况下存在与存在的区别和为什么不创建相同的查询计划有什么区别？
2. 如何让Django ORM生成更有效的查询？

编辑：django模型如下：

  class Album(BaseModel):
    title = models.CharField(max_length=255, blank=False)
    artwork_file = models.ForeignKey(
        S3File, null=True, on_delete=models.CASCADE,
        related_name="album_artwork_file")
    user = models.ForeignKey(settings.AUTH_USER_MODEL,
                             related_name="albums",
                             on_delete=models.CASCADE)
    description = models.TextField(blank=True)
    tags = ArrayField(models.CharField(
        max_length=16), default=default_arr)
    genres = ArrayField(models.CharField(
        max_length=16), default=default_arr)



class Song(BaseModel):
    title = models.CharField(max_length=255, blank=False)
    album = models.ForeignKey(Album,
                              related_name="songs",
                              on_delete=models.CASCADE)
    audio_file = models.ForeignKey(
        S3File, null=True, on_delete=models.CASCADE,
        related_name="song_audio_file")

以下操作不起作用，因为如果在此QuerySet上使用get（），它将引发异常

Album.objects.filter(songs__audio_file__isnull=False).get(pk=1)
Album.MultipleObjectsReturned: get() returned more than one Album 

查询集与DjangoRest ModelViewSet一起使用，其中查询集用于crud操作，并传递给Album Serializer。这需要get（）工作并返回单个值。

class AlbumViewSet(viewsets.ModelViewSet):

    serializer_class = AlbumSerializer

    def get_queryset(self): 

        valid_songs = Song.objects.filter(
            album=OuterRef('pk'),
            audio_file__isnull=False).only('album')

        # Slow query posted above
        return Album.objects.annotate(
            valid_song=Exists(valid_songs)
        ).filter(valid_song=True)

Answer 1

我不确定你为什么要做这些查询。查找至少有一首歌具有音频文件的专辑简称为：

Album.objects.filter(song__audio_file__isnull=False)