django中分别获取父类子类对象

最新推荐文章于 2025-07-09 15:56:22 发布

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本文介绍了一个使用 Django 的 ORM 功能实现继承的例子，通过定义基类和子类来展示如何为不同类型的对象（如文章、照片和链接）提供统一的父类接口。这种设计可以简化查询操作，使得从父类查询子类实例变得更加直观。

[code]
from django.db import models
from django.db.models.query import QuerySet

class ChildQuerySet(QuerySet):
def iterator(self):
for obj in super(ChildQuerySet, self).iterator():
yield obj.get_child_object()

class ChildManager(models.Manager):
def get_query_set(self):
return ChildQuerySet(self.model)

class ParentModel(models.Model):
_child_name = models.CharField(max_length=100, editable=False)

class Meta:
abstract = True

def save(self, *args, **kwargs):
self._child_name = self.get_child_name()
super(ParentModel, self).save(*args, **kwargs)

def get_child_name(self):
if type(self) is self.get_parent_model():
return self._child_name
return self.get_parent_link().related_query_name()

def get_child_object(self):
return getattr(self, self.get_child_name())

def get_parent_link(self):
return self._meta.parents[self.get_parent_model()]

def get_parent_model(self):
raise NotImplementedError

def get_parent_object(self):
return getattr(self, self.get_parent_link().name)

[/code]
用法：
[code]
class Post(ParentModel):
title = models.CharField(max_length=50)

objects = models.Manager()
children = ChildManager()

def __unicode__(self):
return self.title

def get_parent_model(self):
return Post

class Article(Post):
text = models.TextField()

class Photo(Post):
image = models.ImageField(upload_to='photos/')

class Link(Post):
url = models.URLField()

输出:
>>> Post.objects.all()
[<Post: Django>, <Post: Make a Tumblelog>, <Post: Self Portrait>]

>>> Post.children.all()
[<Link: Django>, <Article: Make a Tumblelog>, <Photo: Self Portrait>]
[/code]