python - 当我使用 Manager 时，Django 执行向后关系查询发生最大递归深度超出错误

Question

代码如下:

汽车源类

class CarSource(models.Model):
    status = models.CharField(max_length=1, blank=True, null=True)
    dealer = models.ForeignKey(Dealer, blank=True, null=True, \
                  on_delete=models.SET_NULL, related_name='cars', \
                  verbose_name=u'dealer own this car')
    objects = CarSourceManager()

经销商类

class Dealer(models.Model):
    ...

CarSourceManager 类:

class CarSourceManager(models.Manager):
    ''' Manage query in CarSource, filter data that was processed successfully.'''
    def get_query_set(self):
        return super(self.__class__, self).get_query_set().filter(status='S')

当我执行此操作时:

from ... import Dealer
d = Dealer.objects.get(id = 2)
d.cars.all()

出现如下错误:

File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
File "/...path..of..error..file../apps/car/managers.py", line 9, in all
return super(self.__class__, self).all().filter(status='S')
RuntimeError: maximum recursion depth exceeded while calling a Python object

我覆盖了CarSourceManager中models.Manager的get_query_set，显然，它继续递归地调用它自己。我看了Manager的代码，但无法理解，请帮助我。

Answer 1

查看这个答案:https://stackoverflow.com/a/18208725/1085511

基本上不能用

super(self.__class__, self)

使用

super(CarSourceManager, self)

相反。

相关管理器的 self.__class__ 与 CarSourceManager 不同，因此循环。