Django可选url参数(Django optional url parameters)

Django可选url参数(Django optional url parameters)

我有一个这样的Django网址：

url( r'^project_config/(?P<product>\w+)/(?P<project_id>\w+)/$', 'tool.views.ProjectConfig', name='project_config' ),

和我的views.py：

def ProjectConfig(request, product, project_id=None, template_name='project.html'): ... # do stuff

问题是我希望project_id参数是可选的。 我希望/project_config/ and /project_config/12345abdce/都是同样有效的url模式，所以如果 project_id被传递，那么我可以使用它。 就像现在这样，如果我尝试访问没有project_id参数的url，我会得到一个404。

I have a Django URL like this:

url( r'^project_config/(?P<product>\w+)/(?P<project_id>\w+)/$', 'tool.views.ProjectConfig', name='project_config' ),

views.py:

def ProjectConfig(request, product, project_id=None, template_name='project.html'): ... # do stuff

The problem is that I want the project_id parameter to be optional.

I want /project_config/ and /project_config/12345abdce/ to be equally valid URL patterns, so that if project_id is passed, then I can use it.

As it stands at the moment, I get a 404 when I access the URL without the project_id parameter.

最满意答案

有几种方法

一个是使用一个花哨的正则表达式(?:/(?P<title>[a-zA-Z]+)/)? 制作正则表达式Django URL令牌可选

另一个更容易遵循的方法是拥有符合您需求的多个规则，均指向相同的视图。

urlpatterns = patterns('', url(r'^project_config/$', views.foo), url(r'^project_config/(?P<product>\w+)/$', views.foo), ulr(r'^project_config/(?P<product>\w+)/(?P<project_id>\w+)/$', views.foo), )

请记住，在您看来，您还需要为可选网址参数设置默认值，否则您将收到错误：

def foo(request, optional_parameter=''): # Your code goes here

There are several approaches.

One is to use a non-capturing group in the regex: (?:/(?P<title>[a-zA-Z]+)/)? Making a Regex Django URL Token Optional

Another, easier to follow way is to have multiple rules that matches your needs, all pointing to the same view.

urlpatterns = patterns('', url(r'^project_config/$', views.foo), url(r'^project_config/(?P<product>\w+)/$', views.foo), url(r'^project_config/(?P<product>\w+)/(?P<project_id>\w+)/$', views.foo), )

Keep in mind that in your view you'll also need to set a default for the optional URL parameter, or you'll get an error:

def foo(request, optional_parameter=''): # Your code goes here