CODEWITHSUNDEEP
pythondjangodjango-urls
Houman
Houman

Reputation: 66320

How pass two parameters to url (without GET)?

A while back I was given the advice to not use the GET approach in my urls when using Django as it's cleaner this way.

That works pretty nice with one parameter:

(r'^call/add/(?P<call_id>\d+)/$', call_view),

http://127.0.0.1:8000/call/add/1/

But how could I possibly use the same approach with two parameters?

As I am still learning, please enlighten me about better approaches. Thank you.

Upvotes: 3

Views: 282

Answers (3)

iraycd
iraycd

Reputation: 892

(r'^call/add/(?P<call_id>\d+)/(?P<receiver_id>\d+)/$', call_view),

http://127.0.0.1:8000/call/add/1/903256

and you need to add def call_view(request, call_id, receiver_id): in views.py

or you can you w+ instead of d+ to pass string a a variable

(r'^call/add/(?P<call_id>\d+)/(?P<receiver_name>\w+)/$', call_view),

http://127.0.0.1:8000/call/add/1/Kave

For more info : https://docs.djangoproject.com/en/dev/topics/http/urls/

Upvotes: 2

fdomig
fdomig

Reputation: 4457

You simply can add another on the back like http://127.0.0.1:8000/call/add/1/foo/2. You have to add the second parameter to the regular expression as well like (r'^call/add/(?P<call_id>\d+)/foo/(?P<foo_id>\d+)$', call_view),.

You have to change the controller as well: def call_view(request, call_id, foo_id):

Upvotes: 3

Simeon Visser
Simeon Visser

Reputation: 122326

You can specify multiple parameters as follows:

(r'^call/add/(?P<call_id>\d+)/(?P<other_value>\d+)/$', call_view),

and you view should look like this:

def call_view(request, call_id, other_value):
    # view code here

Upvotes: 2

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