Django 可选 url 参数

问题描述

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.

解决方案

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