Question

在FileView的get_success_URL方法中，我使用了＆＃39; display：redirect＆＃39;作为要重定向的URL。但我收到错误

Reverse for 'redirect' with arguments '()' and keyword arguments '{'pk': 15}' not found. 1 pattern(s) tried: ['display/success/']

请求方法：POST 请求网址：http://127.0.0.1:8000/display/upload/ Django版本：1.10.dev20160512164014 异常类型：NoReverseMatch 例外值：

我错过了什么吗？

display / views.py

class FileView(FormView):

    template_name = 'display/upload.html'
    form_class = FileForm

    def form_valid(self, form):

        file_upload = FileModel(file=self.get_form_kwargs().get('files')['file'])
        file_upload.save()
        self.id = file_upload.id

        return HttpResponseRedirect(self.get_success_url())

    def get_success_url(self):

        return reverse('display:redirect', kwargs={'pk': self.id})

def redirect(request):

    return render(request,"display/success.html")

显示/ urls.py

from django.conf.urls import url
from django.conf.urls.static import static
from django.conf import settings

from . import views

from display.views import FileView

urlpatterns = [
    url(r'^start/', views.initial,name='home'),
    url(r'^upload/',FileView.as_view(),name='upload'),
    url(r'^success/',views.redirect,name='redirect'),
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

Answer 1

看看这一行：

Reverse for 'redirect' with arguments '()' and keyword arguments '{'pk': 15}' not found.

错误消息告诉您错误。您已经传递了一个关键字参数，当它试图反转/找到它没有的网址时，因为您将网址指定为：

url(r'^success/',views.redirect,name='redirect')

哪个不接受任何参数。一种可能性是将你的网址改为：

^success/(?P<pk>\d+)/$

NoReverseMatch在显示/成功/

1 个答案: