表单中捕获的URL参数

Question

我正在使用Userena,我正在尝试捕获URL参数并将它们添加到我的表单中,但我失去了如何做到这一点.

我想在模板中做的是:

<a href="/accounts/signup/freeplan">Free Plan</a><br/>
<a href="/accounts/signup/proplan">Pro Plan</a><br/>
<a href="/accounts/signup/enterpriseplan">Enterprise Plan</a><br/>

然后在我的urls.py中

url(r'^accounts/signup/(?P<planslug>.*)/$','userena.views.signup',{'signup_form':SignupFormExtra}),

然后,理想情况下,我想在forms.py中使用该planslug在配置文件中设置用户计划.

我迷失了如何将捕获的URL参数放入自定义表单中.我可以使用extra_context,是否必须覆盖Userena注册视图？

Answer 1

如果使用基于类的视图,则可以覆盖FormMixin类的def get_form_kwargs()方法.在这里,您可以将所需的任何参数传递给表单类.

在urls.py中:

url(r'^create/something/(?P<foo>.*)/$', MyCreateView.as_view(), name='my_create_view'),

在views.py中:

class MyCreateView(CreateView):
    form_class = MyForm
    model = MyModel

    def get_form_kwargs(self):
        kwargs = super( MyCreateView, self).get_form_kwargs()
        # update the kwargs for the form init method with yours
        kwargs.update(self.kwargs)  # self.kwargs contains all url conf params
        return kwargs

在forms.py中:

class MyForm(forms.ModelForm):

    def __init__(self, foo=None, *args, **kwargs)
        # we explicit define the foo keyword argument, cause otherwise kwargs will 
        # contain it and passes it on to the super class, who fails cause it's not
        # aware of a foo keyword argument.
        super(MyForm, self).__init__(*args, **kwargs)
        print foo  # prints the value of the foo url conf param

希望这可以帮助 :-)