Django optional url parameters

PythonDjangoDjango ViewsDjango Urls

Python Problem Overview


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.

Python Solutions


Solution 1 - Python

There are several approaches.

One is to use a non-capturing group in the regex: (?:/(?P<title>[a-zA-Z]+)/)?
https://stackoverflow.com/questions/2325433/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

Solution 2 - Python

Django > 2.0 version:

The approach is essentially identical with the one given in Yuji 'Tomita' Tomita's Answer. Affected, however, is the syntax:

# URLconf
...

urlpatterns = [
    path(
        'project_config/<product>/',
        views.get_product, 
        name='project_config'
    ),
    path(
        'project_config/<product>/<project_id>/',
        views.get_product,
        name='project_config'
    ),
]


# View (in views.py)
def get_product(request, product, project_id='None'):
    # Output the appropriate product
    ...

Using path() you can also pass extra arguments to a view with the optional argument kwargs that is of type dict. In this case your view would not need a default for the attribute project_id:

    ...
    path(
        'project_config/<product>/',
        views.get_product,
        kwargs={'project_id': None},
        name='project_config'
    ),
    ...

For how this is done in the most recent Django version, see the official docs about URL dispatching.

Solution 3 - Python

You can use nested routes

Django <1.8

urlpatterns = patterns(''
    url(r'^project_config/', include(patterns('',
        url(r'^$', ProjectConfigView.as_view(), name="project_config")
        url(r'^(?P<product>\w+)$', include(patterns('',
            url(r'^$', ProductView.as_view(), name="product"),
            url(r'^(?P<project_id>\w+)$', ProjectDetailView.as_view(), name="project_detail")
        ))),
    ))),
)

Django >=1.8

urlpatterns = [
    url(r'^project_config/', include([
        url(r'^$', ProjectConfigView.as_view(), name="project_config")
        url(r'^(?P<product>\w+)$', include([
            url(r'^$', ProductView.as_view(), name="product"),
            url(r'^(?P<project_id>\w+)$', ProjectDetailView.as_view(), name="project_detail")
        ])),
    ])),
]

This is a lot more DRY (Say you wanted to rename the product kwarg to product_id, you only have to change line 4, and it will affect the below URLs.

Edited for Django 1.8 and above

Solution 4 - Python

Even simpler is to use:

(?P<project_id>\w+|)

The "(a|b)" means a or b, so in your case it would be one or more word characters (\w+) or nothing.

So it would look like:

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

Solution 5 - Python

Thought I'd add a bit to the answer.

If you have multiple URL definitions then you'll have to name each of them separately. So you lose the flexibility when calling reverse since one reverse will expect a parameter while the other won't.

Another way to use regex to accommodate the optional parameter:

r'^project_config/(?P<product>\w+)/((?P<project_id>\w+)/)?$'

Solution 6 - Python

Django = 2.2

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

Solution 7 - Python

Use ? work well, you can check on pythex. Remember to add the parameters *args and **kwargs in the definition of the view methods

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

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionDarwin TechView Question on Stackoverflow
Solution 1 - PythonYuji 'Tomita' TomitaView Answer on Stackoverflow
Solution 2 - PythonjojoView Answer on Stackoverflow
Solution 3 - PythonJacob ValentaView Answer on Stackoverflow
Solution 4 - PythonJuan José BrownView Answer on Stackoverflow
Solution 5 - PythontarequehView Answer on Stackoverflow
Solution 6 - PythonAzizAhmadView Answer on Stackoverflow
Solution 7 - PythonfranciscorodeView Answer on Stackoverflow