How do i pass GET parameters using django urlresolvers reverse

Question

I am using django 1.2 and going from one view to another using the urlresolvers reverse method.

url = reverse(viewOne)

and I want to pass a get parameter, for example

name = 'joe'

so that in the viewOne if I do

def viewOne(request):
    request.GET['name']

I will get

joe

how do I do that ?

Take a look at that answer that gives a rather nice way to do this kind of thing. — sberder
– sberder, Commented Sep 26, 2012 at 3:07
I created a feature request: code.djangoproject.com/ticket/25582 — guettli
– guettli, Commented Oct 21, 2015 at 9:06

Daniel Roseman · Accepted Answer · 2012-03-06 14:27:58Z

164

GET parameters have nothing to do with the URL as returned by reverse. Just add it on at the end:

url = "%s?name=joe" % reverse(viewOne)

answered Mar 6, 2012 at 14:27

Daniel Roseman

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1 Comment

guettli Over a year ago

This works for this example. Since no quoting gets done, this fails if you use values which need to be quoted.

Amir Ali Akbari · Accepted Answer · 2020-09-25 13:58:34Z

84

A safer and more flexible way:

import urllib
from django.urls import reverse


def build_url(*args, **kwargs):
    get = kwargs.pop('get', {})
    url = reverse(*args, **kwargs)
    if get:
        url += '?' + urllib.urlencode(get)
    return url

then you can use build_url:

url = build_url('view-name', get={'name': 'joe'})

which takes same arguments as reverse, but provides an extra keyword argument get where you can put your GET parameters in it as a dictionary.

edited Sep 25, 2020 at 13:58

answered Oct 31, 2012 at 17:01

Amir Ali Akbari

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2 Comments

Dolan Antenucci Over a year ago

Note: Django's django.utils.http.urlencode function might be preferable since it is a "version of Python’s urllib.urlencode() function that can operate on unicode string": docs.djangoproject.com/en/1.10/ref/utils/…

addmoss Over a year ago

I would rename this function to reverse_with_params() or something similar.

clime · Accepted Answer · 2015-10-21 03:57:19Z

22

This is very similar to Amir's solution but handles lists as well.

from django.core.urlresolvers import reverse
from django.http import QueryDict

def build_url(*args, **kwargs):
    params = kwargs.pop('params', {})
    url = reverse(*args, **kwargs)
    if not params: return url

    qdict = QueryDict('', mutable=True)
    for k, v in params.iteritems():
        if type(v) is list: qdict.setlist(k, v)
        else: qdict[k] = v

    return url + '?' + qdict.urlencode()

Example usage:

>>> build_url('member-list', params={'format': 'html', 'sex': ['male', 'female']})
u'/members/?format=html&sex=male&sex=female'

edited Oct 21, 2015 at 3:57

answered Dec 8, 2013 at 21:15

clime

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2 Comments

motatoes Over a year ago

What's the point of having sex=male&sex=female? I think it should be &sex[]=female&sex[]=female (note the additional []), to be a GET request array

Jonas Schäfer Over a year ago

@potatoes The [] are a PHP convention.

Avik Samaddar · Accepted Answer · 2016-07-13 18:05:40Z

-2

Sorry for the delayed correction on this.

While both the answers here handles the required task too well, i think just a simple function to urlencode a dictionary is the most flexible way of doing this:

import urllib

def getify(dic):
    st = ''
    for K, V in dic.items():
        K = urllib.parse.quote(str(K))
        if isinstance(V, list):
            for v in V:
                st += K + '=' + urllib.parse.quote(str(v)) + '&'
        else:
            st += K + '=' + urllib.parse.quote(str(V)) + '&'
    return st.rstrip('&')

edited Jul 13, 2016 at 18:05

answered Nov 13, 2015 at 21:24

Avik Samaddar

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1 Comment

jamesc Over a year ago

This does not urlencode the dictionary - there is no handling of URL reserved characters (e.g. #, ?) or any type of percent encoding: en.wikipedia.org/wiki/Percent-encoding

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