Django注释外键 - 对象的用户特定状态

问题描述

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