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)
创建Child时,pre_save_parent不会触发：
child = models.Child.objects.create(color="red")
这是预期的行为吗？
解决方法
这是一个关于这个的开放票,#9318.
你的解决方法看起来很好.以下是分别于benbest86和alexr在机票上提出的另外两项建议.
>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)
>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)