无ReverseMatch网址Django的

Question

我有一种情况，看起来像这样：

模型

from django.db import models 

from django.core.urlresolvers import reverse_lazy, reverse 

class State(models.Model): 
    short_name = models.CharField('Abbreviation', max_length=2) 
    state = models.SlugField('State Name', max_length=20) 

    def __str__(self): 
     return self.state 

    def get_absolute_url(self): 
     return reverse('state_list', kwargs={'state': self.state}) 

class City(models.Model): 
    state = models.ForeignKey(State) 
    city = models.CharField('City', max_length=100) 

    class Meta: 
     verbose_name = 'City' 
     verbose_name_plural = 'Cities' 

    def __str__(self): 
     return self.city 

class School(models.Model): 
    name = models.CharField(max_length=69, default='') 

    def __str__(self): 
     return self.name 

    class Meta: 
     verbose_name = 'School' 
     verbose_name_plural = 'Schools' 

查看

from django.shortcuts import get_object_or_404, render 
from django.views.generic import ListView, DetailView, CreateView, UpdateView, DeleteView 

from django.core.urlresolvers import reverse_lazy 

from .models import School, City, State 

def reviews_index(request): 
    state_list = State.objects.all() 
    context = {'states': state_list} 
    return render(request, 'reviews/reviews_index.html', context) 

def state_detail(request, state=None): 
    city_list = City.objects.order_by('city') 
    context = {'cities': city_list} 
    return render(request, 'reviews/state_detail.html', context) 

def city_detail(request, state=None, city=None): 
    school_list = School.objects.all() 
    context = {'schools': school_list} 
    return render(request, 'reviews/city_detail.html', context) 

网址

from django.conf.urls import url, include 

from . import views 

app_name = 'reviews' 
urlpatterns = [ 
    url(r'^$', views.reviews_index, name='reviews_index'), 
    url(r'^(?P<state>[a-z]+)/$', views.state_detail, name='state_detail'), 
    url(r'^(?P<state>[a-z]+)/(?P<city>[a-z]+)/$', views.city_detail, name='city_detail'), 

] 

然而，当我尝试从state_detail模板创建一个链接e将city_detail模板，我得到这个错误：

NoReverseMatch at /reviews/alabama/ 
Reverse for 'city_detail' with arguments '('', 'Auburn')' and keyword arguments '{}' not found. 0 pattern(s) tried: [] 

这是我如何在模板链接：

{% block main_content %} 
    <div class="content"> 
     <div class="row"> 
      {% if cities %} 
      {% for city in cities %} 
      <div class="medium-3 column"> 
       <a href="{% url 'city_detail' state.state city.city %}">{{ city.city }}</a> 
      </div> 
      {% endfor %} 
      {% endif %} 
     </div> 
    </div> 
{% endblock %} 

有人可以告诉我，我做错了，帮我解决这个问题。提前致谢。

Answer 1

city_detail url pattern包含state.state，所以你需要在模板上下文中包含状态。

在您看来，您可以使用get_object_or_404来使用slug获取状态。

def state_detail(request, state=None): 
    state = get_object_or_404(State, state=state) 
    city_list = City.objects.filter(state=state).order_by('city') 
    context = {'cities': `city_list`, 'state': state} 
    return render(request, 'reviews/state_detail.html', context) 

请注意，我已经更改了city_list，因此它只显示您正在查看的状态下的城市。

对于实例和slug使用相同的变量名称state并不是一个好主意。重命名其中的一个是个好主意，例如，到state_obj或state_slug。如果你这样做，你必须确保你更新你的网址，视图和模板以保持一致。