django:__ in query lookup不保持querset中的顺序

Question

我按特定顺序拥有ID

>>> album_ids = [24, 15, 25, 19, 11, 26, 27, 28]
>>> albums = Album.objects.filter( id__in=album_ids, published= True )
>>> [album.id for album in albums]
[25, 24, 27, 28, 26, 11, 15, 19]

我需要查询集中的相册作为album_ids中的id.有人请告诉我如何维护订单？或获取专辑中的专辑？

Answer 1

从Djnago 1.8开始,你可以这样做

from django.db.models import Case, When

pk_list = [10, 2, 1]
preserved = Case(*[When(pk=pk, then=pos) for pos, pk in enumerate(pk_list)])
queryset = MyModel.objects.filter(pk__in=pk_list).order_by(preserved)

Answer 2

假设ID列表不是太大,您可以将QS转换为列表并在Python中对其进行排序:

album_list = list(albums)
album_list.sort(key=lambda album: album_ids.index(album.id))

Answer 3

你不能通过ORM在django中做到这一点.但是你自己实现它很简单:

album_ids = [24, 15, 25, 19, 11, 26, 27, 28]
albums = Album.objects.filter(published=True).in_bulk(album_ids) # this gives us a dict by ID
sorted_albums = [albums[id] for id in albums_ids if id in albums]

Answer 4

对于 Django 版本 >= 1.8，请使用以下代码：

from django.db.models import Case, When

field_list = [8, 3, 6, 4]
preserved = Case(*[When(field=field, then=position) for position, field in enumerate(field_list)])
queryset = MyModel.objects.filter(field__in=field_list).order_by(preserved)

这是数据库级别的 PostgreSQL 查询表示：

SELECT *
FROM MyModel
ORDER BY
  CASE
    WHEN id=8 THEN 0
    WHEN id=3 THEN 1
    WHEN id=6 THEN 2
    WHEN id=4 THEN 3
  END;