Question

我正在使用Django 2.0.8和Python 3.5。我写了一个基类，将行为封装在基类中。

保存对象后，应该立即打印一条消息。但是，当我运行应该保存对象的代码时，得到的消息“消息之火”打印了两次-为什么？

这是我的代码：

Restart-Service WinRM

示例用法（省略了导入）

class Likeable(models.Model):
    likes = GenericRelation(Like)


    def action_is_permissible(self, actionable_object, actor):
        ct = ContentType.objects.get_for_model(actionable_object)
        object_id = actionable_object.id
        found_objects = Like.objects.filter(content_type=ct, object_id=object_id, liker=actor)

        return ((len(found_objects) == 0), ct, object_id, found_objects)


    def add_like(self, actionable_object, actor):

        can_add, ct, object_id, found_objects = self.action_is_permissible(actionable_object, actor)

        print(can_add, ct, object_id, found_objects)

        if can_add:
            print('Save Called!')
            # Create like object and save it
            like = self.likes.create(content_type=ct, object_id=object_id, liker=actor)
            like.save()
        else:
            # do nothing
            print('Nothing doing')
            return


    class Meta:
        abstract = True


class Foo(Likeable):
    name = models.CharField(max_length=255,default='')
    objects = models.Manager()

控制台输出

foo = Foo.objects.get(id=1)
p = User.objects.get(id=1)

foo.add_like(foo, p) # <- nasty API calling convention, but I digress

为什么将消息打印两次-当对象仅在数据库中保存一次时？

Answer 1

这里：

like = self.likes.create(content_type=ct, object_id=object_id, liker=actor)
like.save()

QuerySet.create()实际上确实在实例上调用save()-否则该方法将只是模型的初始化程序的无用重复-因此您确实有两次调用。

django模型-save（）方法中的消息打印两次，但对象保存一次

示例用法（省略了导入）

控制台输出

1 个答案: