Question

我在Django中有两个模型，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，至少一首歌曲必须具有音频文件，以便过滤器返回专辑。我正在使用Postgres。

我试图弄清楚如何通过Django QuerySet来执行此逻辑，但是我不确定如何使用where存在而不存在。

以下是我要开始使用的Django orm语句：

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：

在这种情况下，存在与存在之间有什么区别？为什么不创建相同的查询计划？
如何获取Django ORM以生成更有效的查询？

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")

以下操作无效，因为如果在此QuerySet上使用get（），它将抛出异常

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

该查询集与DjangoRest ModelViewSet一起使用，该查询集用于Crud操作，并传递给Album Serializer。这需要get（）起作用并返回单个值。

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

我不确定您为什么要执行这两个查询。查找专辑中至少有一首歌曲带有音频文件的过程表示为：

>>> s = []
>>> with open('f.txt') as file_in:
...     s = file_in.read().split()
... 
>>> s
['Hello', 'Bye', 'Good']

如何在Django中实现“哪里存在”？

1 个答案: