DJango URL反向错误:reversed()的参数必须是一个序列

Question

这是我的urls.py

from django.conf.urls import include, url
from django.contrib import admin
from common.views import HomeView, LoadingSchoolView, ProcessSchoolView

urlpatterns = [
    url(r'^$', HomeView.as_view(), name='Index'),
    url(r'^admin/', include(admin.site.urls)),
    url(r'^member/', include('member.urls', namespace='member')),
    url(r'^common/', include('common.urls', namespace='common')),

在我的common/urls.py中

from django.conf.urls import url
from .views import QuerySchoolView

urlpatterns = {
    url(r'^querySchool/(?P<q>[a-z]*)$', QuerySchoolView.as_view(), name='querySchool'),
}

现在,当我这样做

{% url 'common:querySchool' %}, 

它给了我一个 TypeError

TypeError at /member/register/learner
argument to reversed() must be a sequence
Request Method: GET
Request URL:    http://127.0.0.1:8000/member/register/learner
Django Version: 1.8.2
Exception Type: TypeError
Exception Value:    
argument to reversed() must be a sequence
Exception Location:       /Users/enzii/python_env/django18/lib/python3.4/site-packages/django/core/urlresolvers.py in _populate, line 285
Python Executable:  /Users/enzii/python_env/django18/bin/python3
Python Version: 3.4.3

这是我的观点

class QuerySchoolView(View):

    def get(self, request, q=""):

        universities = University.objects.filter(Q(name__contains=q) |
                                  Q(pinyin__contains=q) |
                                  Q(province__contains=q) |
                                  Q(country__contains=q))[:4]

        returnObj = [{'unvis-name': u.name, 'country': u.country, 'province': u.province} for u in  universities]

        return HttpResponse(returnObj)

我有什么问题{% url %}？

Answer 1

您没有名为"index"的URL,您有一个名为"Index"的URL.

在您的common/urls中,您使用的是{}代替[]包装模式.

将来,请分别发布每个问题.