问题描述
I have a Django URL like this:
url(
r'^project_config/(?P<product>w+)/(?P<project_id>w+)/$',
'tool.views.ProjectConfig',
name='project_config'
),
views.py:
def ProjectConfig(request, product, project_id=None, template_name='project.html'):
...
# do stuff
The problem is that I want the project_id parameter to be optional.
I want /project_config/ and /project_config/12345abdce/ to be equally valid URL patterns, so that if project_id is passed, then I can use it.
As it stands at the moment, I get a 404 when I access the URL without the project_id parameter.
There are several approaches.
One is to use a non-capturing group in the regex: (?:/(?P<title>[a-zA-Z]+)/)?
Making a Regex Django URL Token Optional
Another, easier to follow way is to have multiple rules that matches your needs, all pointing to the same view.
urlpatterns = patterns('',
url(r'^project_config/$', 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
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