我有一个带有模型管理器的简单模型:
class CompanyReviewManager(models.Manager):
def get_votes_for_company(self, company):
try:
return CompanyReview.objects.filter(user = user).count()
except ObjectDoesNotExist:
return None
def get_rating_for_field(self, installer, field):
try:
return CompanyReview.objects.filter(user = user).aggregate(Avg(field))
except ObjectDoesNotExist:
return None
class CompanyReview(models.Model):
user = models.ForeignKey(settings.AUTH_USER_MODEL)
satisfaction = models.PositiveSmallIntegerField(blank = True, null = True,)
comments = models.TextField(blank = True, null = True,)
objects = CompanyReviewManager()
def save(self, *args, **kwargs):
obj = super(InstallerReview, self).save(*args, **kwargs)
return obj当我现在试图在Django shell中保存一个对象时,该对象将被保存,但不会返回任何内容。为什么?
In [1]: company_obj = InstallerReview()
In [2]: company_obj.user = CompanyUser.objects.all()[2]
In [3]: obj = company_obj.save()
In [4]: obj
Out[4]:
In [5]: company_obj
Out[5]: <CompanyReview: AdminCompany>为什么第三步没有出错?
发布于 2013-05-10 15:31:53
因为超类save方法不返回任何内容。它不需要:self正在被保存,没有必要返回其他东西并将其命名为obj。
您可以从子类self方法返回save,但没有太多意义。通常,在Python中,如果函数更改了对象,它们就不会返回更改的对象:与list sort()方法进行比较。
https://stackoverflow.com/questions/16485651
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