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Python django模型实例don';保存()后不会更改_Python_Django_Django Models - Fatal编程技术网

Python django模型实例don';保存()后不会更改

Python django模型实例don';保存()后不会更改,python,django,django-models,Python,Django,Django Models,这里,公司是一个模型实例,名称和证书在保存()后不会更改 然而 def post(self, request, *args, **kwargs): self.company.name = request.POST['company_name'] self.company.certification = request.POST['company_pic_url'] self.company.save() return http.HttpResponseRedire

这里,公司是一个模型实例,名称和证书在保存()后不会更改

然而

def post(self, request, *args, **kwargs):
    self.company.name = request.POST['company_name']
    self.company.certification = request.POST['company_pic_url']
    self.company.save()
    return http.HttpResponseRedirect('/company')
它工作得很好
有人能解释吗?谢谢

如何定义自己的公司?如果它是一个普通的Django模型实例,那么这两个代码段没有理由有不同的结果。
self.company=company.objects.get(id=1)
Company也是一个普通的Django模型。您是否确保调用了post?是否确实在呈现视图后提交了事务?您在代码中实际在哪里设置了
self.Company=Company.objects.get(id=1)
?例如,如果
self.company
是一个属性,则该代码将无法工作,因为每次执行
self.company
时都会重新获取该代码。
def post(self, request, *args, **kwargs):
    company = self.company
    company.name = request.POST['company_name']
    company.certification = request.POST['company_pic_url']
    company.save()
    return http.HttpResponseRedirect('/company')