如何在 Django 自定义模型管理器中访问相关模型?

时间:2021-03-24 13:49:00

标签: python django django-models django-rest-framework

我有以下三种型号, models.py

class Activity(models.Model):
    # ...

    objects = models.Manager()
    activity = ActivityManager()

    kind = models.CharField(max_length=10, choices=KIND_CHOICES)

    # ...

class ActivtyA(models.Model):
    # ...

    activity = models.OneToOneField(
        Activity,
        on_delete=models.CASCADE,
        related_name='A',
    )

    # ...

class ActivtyB(models.Model):
    # ...

    activity = models.OneToOneField(
        Activity,
        on_delete=models.CASCADE,
        related_name='B',
    )

    # ...

和模特经理, managers.py

class ActivityManager(models.Manager):
    def create_activity(self, validated_data):
        # ...

        data = validated_data.pop(kind)
        act = self.create(**validated_data)
        if act.kind == 'A':
            # Create ActivityA(act, **data)
        elif act.kind == 'B':
            # Create ActivityB(act, **data)

         # ... 
    # ... 

在模型管理器的 create_activity 方法中,我想根据 Activty 创建 ActivityAActivtyBActivity.kind。如果我在管理器中导入这些类,则会导致循环导入错误
如何在管理器中访问 ActivityAActivtyB

我试图通过使用信号来做到这一点,但没有成功。

@receiver(post_save, sender=Activity)
def create_activity_details(sender, instance, using, **kwargs):
    if instance.kind == 'A':
        ActivityA.objects.create(activity=instance, data=????) # Need data to create this object
    elif instance.kind == 'A':
        ActivityB.objects.create(activity=instance, data=????)

1 个答案:

答案 0 :(得分:1)

所以,我回到了老学校并这样做是为了让它发挥作用,

data = validated_data.pop(kind)
act = self.create(**validated_data)
if act.kind == 'A':
    from activity.models import ActivityA  # Prevent the circular import
    ActivityA.objects.create(activity=act, **data)
elif act.kind == 'B':
    from activity.models import ActivityB
    ActivityB.objects.create(activity=act, **data)

它有效,但看起来不干净。还有其他更好的解决方案吗?


更新:

因此,使用 django.apps.apps.get_model 就是答案。感谢 @djvj 为我指明了正确的方向。
医生说:

<块引用>

apps.get_model(app_label, model_name, require_ready=True)
返回具有给定 app_labelmodel_name 的模型。作为一种快捷方式,此方法还接受 app_label.model_name 形式的单个参数。 model_name 不区分大小写。

示例:

from django.apps import apps

class ActivityManager(models.Manager):
def create_activity(self, validated_data):
    # ...

    data = validated_data.pop(kind)
    act = self.create(**validated_data)
    if act.kind == 'A':
        model = apps.get_model(app_label='activity', model_name='ActivityA')
        model.objects.create(activity=act, **data)
    elif act.kind == 'B':
        model = apps.get_model(app_label='activity', model_name='ActivityB')
        model.objects.create(activity=act, **data)

     # ... 
# ... 

apps.get_model(app_label='activity', model_name='ActivityA') 可以简单地写成

apps.get_model('activity.ActivityA')