假设我有一个模型,Box一个GenericForeignKey指向任何一个Apple实例或Chocolate实例.Apple而Chocolate反过来,有ForeignKeys来Farm和Factory,分别.我想显示一个Boxes 列表,我需要访问Farm它Factory.如何在尽可能少的数据库查询中执行此操作？
最小的说明性示例:
class Farm(Model):
...
class Apple(Model):
farm = ForeignKey(Farm)
...
class Factory(Model):
...
class Chocolate(Model):
factory = ForeignKey(Factory)
...
class Box(Model)
content_type = ForeignKey(ContentType)
object_id = PositiveIntegerField()
content_object = GenericForeignKey('content_type', 'object_id')
...
def __unicode__(self):
if self.content_type == ContentType.objects.get_for_model(Apple):
apple = self.content_object
return "Apple {} from Farm {}".format(apple, apple.farm)
elif self.content_type == ContentType.objects.get_for_model(Chocolate):
chocolate = self.content_object
return …我有一种情况,其中特定类的大量对象被迭代,并且它们花费大量时间进行处理,因为我不能使用预先选择数据select_related.
有问题的课程如下所示
from django.contrib.contenttypes.models import ContentType
from django.db import models
class Offer(models.Model):
...
object_id = models.PositiveIntegerField(db_index = True)
content_type = models.ForeignKey(ContentType, db_index = True)
content_object = generic.GenericForeignKey('content_type', 'object_id')
...
我尝试使用如下所示的select_related,但它显然不起作用
offerList = Offer.objects.select_related('content_type', "content_object"
).filter(content_type=ContentType.objects.get_for_model(SomeObject),
object_id=someobject.id)
那么,我如何select_related在django中使用GenericForeignKey？