如何为所有其他在django中找不到的url编写模式? [英] How to write pattern for all other urls that basically goes for not found in django?
问题描述
我尝试过这个,
url('',views.notfound,name ='notfound')
url ('', views.notfound, name='notfound')
但似乎无法正常工作,例如,我定义了另一个网址模式,
But seems it doesn't work properly, for example, I have another url pattern define,
url(r'^ login / $',views.login,name ='login'),
url(r'^login/$', views.login, name='login'),
所以,如果我去 http://example.com/login/
,这行得通,但是如果我去 http:// example.com/login/?help=1
,则它属于未找到类别。我该如何处理?
So, if I go for http://example.com/login/
, this works, but if i go for http://example.com/login/?help=1
, then it falls into notfound category. How can I handle that?
推荐答案
如果您只想创建一个找不到URL时显示的页面(即抛出Http404异常),则可以创建一个名称为 404.html
的模板,并且该模板将在找不到URL时显示。
If you just want to create a page that is displayed when a URL is not found (i.e. a Http404 exception is thrown) you can create a template with the name 404.html
, and that template will be displayed any time a URL is not found.
或者,如果要定义自定义404处理程序视图,则可以在中定义
文件。然后,只需创建您的 handler404 = views.notfound
urls.py notfound
视图,并且只要抛出404错误(并且 DEBUG = False
)
Or if you want to define a custom 404 handler view you can define handler404 = views.notfound
in your urls.py
file. Then just create your notfound
view, and that view will be used whenever a 404 error is thrown (and DEBUG = False
)
这是捕获任何无法识别的URL并显示友好的404页面的更好方法。
This is a better way to catch any urls that are not recognized and display a friendly 404 page.
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