Question

所以我正在制作电视节目的情节跟踪器，目前正在制作“添加节目”形式。当我去localhost：8000 / add时，它说没有找到。我可能做错了什么？

views.py

class ShowCreate(CreateView):
    model = Show
    fields = ['title', 'description', 'episode', 'season']

的index.html

<a href="{% url 'show:show-add' %}">Add</a>

urls.py

url(r'^add/$', views.ShowCreate.as_view(), name='show-add'),

show_form.html

{% extends 'show/base.html' %}
{% block title %}Add Show{% endblock %}

{% block content %} 

<form action="" method="post">
{% csrf_token %}
<h1>Add show</h1>
{% include 'show/form-template.html' %}
<button>Submit</button>
</form>

{% endblock %}

form_template.html

{% for field in form %}
<div class="form">
    <h3>{{field.label_tag}}</h3>
    <div class="field">{{field}}</div>
</div><!--form-->
{% endfor %}

这是urlpatterns中的所有内容

urlpatterns = [

  # index
  url(r'^$', views.IndexView.as_view(), name='index'),

  # show detail
  url(r'^(?P<show>[\w.@+-]+)/$', views.ShowDetail.as_view(), name='show-detail'),

  # form to add show
  url(r'^add/$', views.ShowCreate.as_view(), name='show-add'),

  # delete show

  # signup

  # login

  # logout

]

错误讯息：

Page not found (404)
Request Method: GET
Request URL:    http://localhost:8000/add/
Raised by:  show.views.ShowDetail

main urls.py

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^', include('show.urls')),
]

Django：URL无法识别？不知道我做错了什么

0 个答案: