本文介绍了如何在Django中显示自定义的404.html页面的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想在最终用户输入错误的URL时显示自定义的404错误页面,我已经尝试过,但我只得到Django默认的404页面。我使用Python(2.7.5),Django(1.5.4)我的代码
urls.py
来自django.conf.urls导入模式的包含来自mysite导入视图的url
handler404 = views.error404
urlpatterns = patterns('',
url(r'^ $','mysite.views.home',name ='home'),
)
views.py
from django.http import HttpResponse
from django $ s
返回HttpResponse(欢迎来到我的世界)
def error404(request):
template = loader.get_template('404.html')
context = Context({'message':'All:%s'%request,})
返回HttpResponse(content = template.render(context),content_type ='text / html; charset = utf-8',status = 404)
404.html页面在templates目录下。如何处理这个..?
解决方案
自定义URL处理程序不适用于 DEBUG =真
。设置 DEBUG = False
,它将工作。 (您还需要设置 ALLOWED_HOSTS = ['127.0.0.1']
)
I want to display custom 404 error page when end user enters wrong url,I have tried but i am getting only Django default 404 page.I am using Python(2.7.5),Django(1.5.4)
My Code
urls.py
from django.conf.urls import patterns, include, url
from mysite import views
handler404 = views.error404
urlpatterns = patterns('',
url(r'^$', 'mysite.views.home', name='home'),
)
views.py
from django.http import HttpResponse
from django.shortcuts import render
from django.template import Context, loader
def home(request):
return HttpResponse("welcome to my world")
def error404(request):
template = loader.get_template('404.html')
context = Context({'message': 'All: %s' % request,})
return HttpResponse(content=template.render(context), content_type='text/html; charset=utf-8', status=404)
I have placed my 404.html page in templates directory.How to handle this..?
解决方案
Custom URL handlers don't work with DEBUG=True
. Set DEBUG=False
and it will work. (You'll also need to set ALLOWED_HOSTS=['127.0.0.1']
)
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