def clean(self):
cleaned_data = self.cleaned_data
current_pass = cleaned_data['current_pass']
new_pass = cleaned_data['new_pass']
new_pass2 = cleaned_data['new_pass2']
if current_pass or new_pass or new_pass2:
if not current_pass:
raise forms.ValidationError("- You must enter your current password.")
if not new_pass:
raise forms.ValidationError("- You must enter a new password.")
if not new_pass2:
raise forms.ValidationError("- You must re-confirm your new password.")
return cleaned_data
现在,我提出我的错误。但这意味着其他错误不会弹出。它会在我提出第一个函数时结束函数。如果我想要全部3个错误怎么办?在Django窗体中,如何将错误添加到队列中?
所以,我应该把错误列表中,并提出名单? – TIMEX 2010-12-13 23:51:17
不,您应该引发_one_错误,您应根据发生的故障创建错误。 – katrielalex 2010-12-14 01:46:19