Django URL模板匹配(除模式外的所有内容) [英] Django URL template match (everything except pattern)

问题描述

我需要一个Django regex，它实际上将对url路由器起作用，以执行以下操作:

I need a django regex that will actually work for the url router to do the following:

匹配路由中不包含"/api"的所有内容.

Match everything that does not contain "/api" in the route.

以下内容不起作用，因为Django无法反转(?！

The following does not work because django can't reverse (?!

r'^(?!api)

推荐答案

通常的解决方法是对路由声明进行排序，以使/api 路线，即:

The usual way to go about this would be to order the route declarations so that the catch-all route is shadowed by the /api route, i.e.:

urlpatterns = patterns('', 
    url(r'^api/', include('api.urls')),
    url(r'^other/', 'views.other', name='other'),
    url(r'^.*$', 'views.catchall', name='catch-all'), 
)

或者，如果由于某些原因您确实需要跳过某些路由，但不能使用Django支持的一组正则表达式来做到这一点，则可以定义一个自定义模式匹配器类:

Alternatively, if for some reason you really need to skip some routes but cannot do it with the set of regexes supported by Django, you could define a custom pattern matcher class:

from django.core.urlresolvers import RegexURLPattern 

class NoAPIPattern(RegexURLPattern):
    def resolve(self, path):
        if not path.startswith('api'):
            return super(NoAPIPattern, self).resolve(path)

urlpatterns = patterns('',
    url(r'^other/', 'views.other', name='other'),
    NoAPIPattern(r'^.*$', 'views.catchall', name='catch-all'),
)

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