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?
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))
从Django 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)
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]
你可以使用extra QuerySet修饰符通过ORM在Django中完成
>>> album_ids = [24, 15, 25, 19, 11, 26, 27, 28]
>>> albums = Album.objects.filter( id__in=album_ids, published= True
).extra(select={'manual': 'FIELD(id,%s)' % ','.join(map(str, album_ids))},
order_by=['manual'])