Question

当我调出127.0.0.1:8000时，显示的当前页面为something.html模板。

我需要在首次启动时使其显示为index.html，然后当我点击其他部分时，它应该转到127.0.0.1:8000/something.html（或127.0.0.1:8000/myapp/something。 HTML）。

实现这一目标的结构是什么？
我经常收到错误消息：The current URL didn't match any of these.

目前，我的结构是

project
     ---myapp
          ---admin.py
          ---models.py
          ---url.py
          ---views.py
     ---static
     ---templates
          ---myapp
             ---html files
     ---mysite
          ---settings.py
          --- url.py

在我的settings.py

下

STATIC_URL = '/static/'
STATIC_ROOT = os.path.join(BASE_DIR, 'static')

mysettings / url.py

   urlpatterns = [
        # Examples:
        url(r'^$', 'mysite.views.home', name='home'),
        url(r'^myapp/', include('myapp.urls')),
        url(r'^admin/', include(admin.site.urls)),
    ]

的mysite / urls.py

from django.conf.urls import include, url
from django.contrib import admin
from django.conf.urls import patterns
from django.views.generic import TemplateView


urlpatterns = [
    # Examples:
    url(r'^', include('myapp.urls')),
    url(r'^admin/', include(admin.site.urls)),
]


admin.site.site_header = 'Admin'

的myapp / url.py

from . import views
from django.conf.urls import patterns, include, url
from django.views.generic import TemplateView


urlpatterns = patterns('',
    #brings up something.html when we comment it out. No module found if included.
    #url(r'^$', views.home, name='home'),
    url(r'^$', views.post_list, name='post_list'),
)

Answer 1

在myapp

mysite/urls.py

from django.conf.urls import include, url
from django.contrib import admin
from django.conf import settings

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

]

现在，从127.0.0.1:8000开始的所有网址都会检查是否有可以处理请求的视图。我建议您阅读本文以了解Django URL调度程序的工作原理：[Django文档 - URL调度程序]（https://docs.djangoproject.com/en/1.8/topics/http/urls/）。 2.在myapp.urls中添加新路线：

from django.conf.urls import url, patterns
from . import views

urlpatterns = patterns('',
    url(r'^$', views.home, name='home'),
    url(r'^something$', views.something, name='something'),
    url(r'^posts$', views.post_list, name='post_list'),
)

现在：
127.0.0.1:8000/将执行views.home的代码 127.0.0.1:8000/something将执行views.something的代码 127.0.0.1:8000/posts将执行views.post_list的代码我们现在定义这些视图

3：在myapp.views中定义视图：

from django.shortcuts import render

def home(request):
    """
    Return home page
    """
    return render(request, 'myapp/home.html')

def something(request):
    """
    Return something page
    """
    return render(request, 'myapp/something.html')

def post_list(request):
    """
    Return something page
    """
    # do what you want

在myapp/templates/myapp/中添加模板。添加home.html和something.html。

现在忘了，`.html

Answer 2

您创建了url（附加了view）。
在view中，您可以呈现所需的任何html页面。

如果您使用基于功能的视图，则'mysite.views.home'可能如下所示：

def home(request):
    ...
    return render(request, 'path/to/index.html')

等等。

这是基础，所以我不会谈论这么多。您可以找到关于将网址映射到视图there的好教程。

如何配置Django视图和URL以呈现特定模板

2 个答案: