Django:基本模型信号处理程序不会触发

Question

在以下示例代码中:

from django.db import models
from django.db.models.signals import pre_save

# Create your models here.
class Parent(models.Model):
    name = models.CharField(max_length=64)

    def save(self, **kwargs):
        print "Parent save..."
        super(Parent, self).save(**kwargs)

def pre_save_parent(**kwargs):
    print "pre_save_parent"
pre_save.connect(pre_save_parent, Parent)

class Child(Parent):
    color = models.CharField(max_length=64)

    def save(self, **kwargs):
        print "Child save..."
        super(Child, self).save(**kwargs)

def pre_save_child(**kwargs):
    print "pre_save_child"
pre_save.connect(pre_save_child, Child)

pre_save_parent 我创建一个Child时不会触发:

child = models.Child.objects.create(color="red")

这是预期的行为吗？

Answer 1

有关于此的公开票,#9318.

你的解决方法看起来很好.以下是benbest86和alexr分别在票上建议的其他两个.

1. 听取子类信号,并在那里发送父信号.

   def call_parent_pre_save(sender, instance, created, **kwargs):
       pre_save.send(sender=Parent, instance=Parent.objects.get(id=instance.id), created=created, **kwargs)
   pre_save.connect(call_parent_pre_save, sender=Child)
   

2. 连接信号时不要指定发送方,然后检查父类的子类.

   def pre_save_parent(sender, **kwargs):
       if not isinstance(instance, Parent):
           return
       #do normal signal stuff here
       print "pre_save_parent"
   
   pre_save.connect(pre_save_parent)