如何构建多个提交按钮 django 表单？

Question

我有一个表单，其中包含一个电子邮件输入和两个提交按钮，用于订阅和取消订阅时事通讯：

<form action="" method="post">
{{ form_newsletter }}
<input type="submit" name="newsletter_sub" value="Subscribe" />
<input type="submit" name="newsletter_unsub" value="Unsubscribe" />
</form>

我也有上课形式：

class NewsletterForm(forms.ModelForm):
    class Meta:
        model = Newsletter
        fields = ('email',)

我必须编写自己的 clean_email 方法，并且我需要知道表单是通过哪个按钮提交的。但是提交按钮的值不在self.cleaned_data 字典中。 否则我可以获取按钮的值吗？

Answer 1

例如：

if 'newsletter_sub' in request.POST:
    # do subscribe
elif 'newsletter_unsub' in request.POST:
    # do unsubscribe

Answer 2

您可以在clean_email 方法中使用self.data 在验证之前访问POST 数据。它应该包含一个名为 newsletter_sub 或 newsletter_unsub 的键，具体取决于按下的按钮。

# in the context of a django.forms form

def clean(self):
    if 'newsletter_sub' in self.data:
        # do subscribe
    elif 'newsletter_unsub' in self.data:
        # do unsubscribe

Answer 3

你也可以这样做，

 <form method='POST'>
    {{form1.as_p}}
    <button type="submit" name="btnform1">Save Changes</button>
    </form>
    <form method='POST'>
    {{form2.as_p}}
    <button type="submit" name="btnform2">Save Changes</button>
    </form>

代码

if request.method=='POST' and 'btnform1' in request.POST:
    do something...
if request.method=='POST' and 'btnform2' in request.POST:
    do something...

Answer 4

同一个视图的一个网址！ 就这样！

urls.py

url(r'^$', views.landing.as_view(), name = 'landing'),

views.py

class landing(View):
        template_name = '/home.html'
        form_class1 = forms.pynamehere1
        form_class2 = forms.pynamehere2
            def get(self, request):
                form1 = self.form_class1(None)
                form2 = self.form_class2(None)
                return render(request, self.template_name, { 'register':form1, 'login':form2,})

             def post(self, request):
                 if request.method=='POST' and 'htmlsubmitbutton1' in request.POST:
                        ## do what ever you want to do for first function ####
                 if request.method=='POST' and 'htmlsubmitbutton2' in request.POST:
                         ## do what ever you want to do for second function ####
                        ## return def post###  
                 return render(request, self.template_name, {'form':form,})

/home.html
    <!-- #### form 1 #### -->
    <form action="" method="POST" >
      {% csrf_token %}
      {{ register.as_p }}
    <button type="submit" name="htmlsubmitbutton1">Login</button>
    </form>
    <!--#### form 2 #### -->
    <form action="" method="POST" >
      {% csrf_token %}
      {{ login.as_p }}
    <button type="submit" name="htmlsubmitbutton2">Login</button>
    </form>

Answer 5

现在这是一个老问题了，不过我遇到了同样的问题并找到了一个适合我的解决方案：我写了 MultiRedirectMixin。

from django.http import HttpResponseRedirect

class MultiRedirectMixin(object):
    """
    A mixin that supports submit-specific success redirection.
     Either specify one success_url, or provide dict with names of 
     submit actions given in template as keys
     Example: 
       In template:
         <input type="submit" name="create_new" value="Create"/>
         <input type="submit" name="delete" value="Delete"/>
       View:
         MyMultiSubmitView(MultiRedirectMixin, forms.FormView):
             success_urls = {"create_new": reverse_lazy('create'),
                               "delete": reverse_lazy('delete')}
    """
    success_urls = {}  

    def form_valid(self, form):
        """ Form is valid: Pick the url and redirect.
        """

        for name in self.success_urls:
            if name in form.data:
                self.success_url = self.success_urls[name]
                break

        return HttpResponseRedirect(self.get_success_url())

    def get_success_url(self):
        """
        Returns the supplied success URL.
        """
        if self.success_url:
            # Forcing possible reverse_lazy evaluation
            url = force_text(self.success_url)
        else:
            raise ImproperlyConfigured(
                _("No URL to redirect to. Provide a success_url."))
        return url

Answer 6

我知道这已经过时了，但至少可以说，有些答案很简短，而且它们没有解决表单不是 django 表单的常见情况。

此解决方案的灵感来自 this blog post。它依赖于使用从 django.views.generic.edit.FormMixin 派生的视图类，例如CreateView、UpdateView 或 DeleteView。这些提供了 get_success_url 方法，该方法在 request

中公开按钮名称

html

<html>
    <body>
        <form method="post">
            <div>
                <label> <input type="radio" name="select-type" value="A">Type A</label>
            </div>
            <div>
                <label> <input type="radio" name="select-type" value="B">Type B</label>
            </div>
            <div>
                <input type="submit" value="Use selected">
            </div>
            <div>
                <input type="submit" name="no-selection" value="None of the above">
            </div>
        </form>
    </body>
</html>

views.py

from django.views.generic import UpdateView

class GetType(UpdateView):
    def get(self, request):
        return render(request, 'get_type.html', {})

    def post(self, request):
        button = self.get_success_url()
        print(button)

    def get_success_url(self):
        if 'no-selection' in self.request.POST:
            return 'none selected'
        return ''