我有一个带有简单验证码的python / django应用程序。 根据某种算法出现了简单的验证码。
有两个几乎相同的构造函数。区别仅在于继承。 如何避免代码重复?
class PostCaptchaForm(PostForm):
captcha = CaptchaField()
def __init__(self, *args, **kwargs):
self.request = kwargs['request']
del kwargs['request']
super(PostCaptchaForm, self).__init__(*args, **kwargs)
def clean(self):
cleaned_data = super(PostCaptchaForm, self).clean()
success = self.is_valid()
utils.update_captcha_access(self.request, success)
if success:
return cleaned_data
else:
raise forms.ValidationError("captcha validation failed")
class ThreadCaptchaForm(ThreadForm):
captcha = CaptchaField()
def __init__(self, *args, **kwargs):
self.request = kwargs.pop('request')
super(ThreadCaptchaForm, self).__init__(*args, **kwargs)
def clean(self):
cleaned_data = super(ThreadCaptchaForm, self).clean()
success = self.is_valid()
utils.update_captcha_access(self.request, success)
if success:
return cleaned_data
else:
raise forms.ValidationError("captcha validation failed")
答案 0 :(得分:3)
我使用多重继承,使用基本验证码形式作为mixin,与ThreadForm或PostForm一起使用
class CaptchaFormBase(forms.Form): # or forms.ModelForm, if appropriate
captcha = CaptchaField()
def __init__(self, *args, **kwargs):
self.request = kwargs.pop('request')
# this is the same as assigning to self request then deleting the key
super(CaptchaFormBase, self).__init__(*args, **kwargs)
def clean(self):
cleaned_data = super(CaptchaFormBase, self).clean()
success = self.is_valid()
utils.update_captcha_access(self.request, success)
if not success:
# this will put things in non_field_errors, you may want to put it in self.errors['captcha']
raise forms.ValidationError("Captcha validation failed")
# always make it easy to see that you're returning cleaned_data
return self.cleaned_data
class PostCaptchaForm(PostForm, CaptchaFormBase):
pass
class ThreadCaptchaForm(ThreadForm, CaptchaFormBase):
pass