如何解决Django注释外键 - 对象的用户特定状态
有 Widget
和 UserStatus
模型。
挑战:获取小部件的查询集,用来自用户和小部件的统一状态进行注释。
逻辑:如果有用户状态,则使用该状态,否则使用小部件的状态。
https://gist.github.com/Lucianovici/53c00ec62f631580dac774cfa4a1578b
class WidgetManager(models.Manager):
def with_user_status(self,user: User = None):
"""This approach duplicates the same widget for each UserStatus"""
return self.annotate(
unified_status_id=Case(
When(user_status__user=user,then='user_status__status_id'),default=F('status_id'),output_field=models.IntegerField(),)
).order_by('unified_status_id')
class Widget(models.Model):
STATUS_NOK = 1
STATUS_OK = 2
STATUS_NONE = 3
name = models.CharField(max_length=40)
status_id = models.IntegerField(default=STATUS_NONE)
objects = WidgetManager()
def __str__(self):
return f'{self.name} - status: {self.status_id}'
class UserStatus(models.Model):
user = models.ForeignKey(to=User,on_delete=models.CASCADE)
widget = models.ForeignKey(to=Widget,on_delete=models.CASCADE,related_name='user_status')
status_id = models.IntegerField(default=Widget.STATUS_NONE)
def __str__(self):
return f'{self.widget.name} - status: {self.status_id}'
让我们试一试。
u1 = User.objects.filter(username='user1').first() or User.objects.create_user('user1')
u2 = User.objects.filter(username='user2').first() or User.objects.create_user('user2')
w1 = Widget.objects.create(name='Widget1',status_id=Widget.STATUS_NONE)
w2 = Widget.objects.create(name='Widget2',status_id=Widget.STATUS_OK)
w3 = Widget.objects.create(name='Widget3',status_id=Widget.STATUS_NOK)
UserStatus.objects.create(user=u1,widget=w1,status_id=Widget.STATUS_NOK)
UserStatus.objects.create(user=u2,status_id=Widget.STATUS_OK)
qs = Widget.objects.all()
print(f'All widgets: {qs}')
qs = Widget.objects.with_user_status(user=u1)
print(f'Widgets with user status {[(w.name,w.unified_status_id) for w in qs]}')
我明白了:
All widgets: <QuerySet [<Widget: Widget1 - status: 3>,<Widget: Widget2 - status: 2>,<Widget: Widget3 - status: 1>]>
Widgets with user status [('Widget1',1),('Widget3',('Widget2',2),('Widget1',3)]
预期结果:
Widgets with user status [('Widget1',2)]
我也在 QuerySet 的末尾尝试了 .distinct()
,但没有任何运气。
谢谢!
解决方法
好吧,我有一个解决方案,但我并不为此感到自豪:) 我认为它可以做得比这更好。
class WidgetManager(models.Manager):
def with_user_status(self,user: User = None):
return self.annotate(
user_id=F('user_status__user'),unified_status_id=Case(
When(user_status__user=user,then='user_status__status_id'),default=F('status_id'),output_field=models.IntegerField(),)
).exclude(
Q(user_id__isnull=False) & ~Q(user_id=user.id)
).order_by('unified_status_id')
我所做的是exclude
所有其他具有不同用户状态的小部件。
qs = Widget.objects.with_user_status(user=u1)
print(f'Widgets with user status {[(w.name,w.unified_status_id,w.user_id) for w in qs]}')
输出:
Widgets with user status [('Widget1',1,428),('Widget3',None),('Widget2',2,None)]
其中 428
是我的 u1
用户 ID。
SQL:
SELECT "core_widget"."id","core_widget"."name","core_widget"."status_id","core_userstatus"."user_id" AS "user_id",CASE
WHEN "core_userstatus"."user_id" = 428 THEN "core_userstatus"."status_id"
ELSE "core_widget"."status_id" END AS "unified_status_id"
FROM "core_widget"
LEFT OUTER JOIN "core_userstatus" ON ("core_widget"."id" = "core_userstatus"."widget_id")
WHERE NOT ("core_userstatus"."user_id" IS NOT NULL AND NOT ("core_userstatus"."user_id" = 428))
ORDER BY "unified_status_id" ASC
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