Django query_transform templatetag

templatetags.py

from django import template

register = template.Library()

@register.simple_tag(takes_context=True)
def query_transform(context, **kwargs):
    '''
    Returns the URL-encoded querystring for the current page,
    updating the params with the key/value pairs passed to the tag.
    
    E.g: given the querystring ?foo=1&bar=2
    {% query_transform bar=3 %} outputs ?foo=1&bar=3
    {% query_transform foo='baz' %} outputs ?foo=baz&bar=2
    {% query_transform foo='one' bar='two' baz=99 %} outputs ?foo=one&bar=two&baz=99
    
    A RequestContext is required for access to the current querystring.
    '''
    query = context['request'].GET.copy()
    for k, v in kwargs.items():
        query[k] = v
    return query.urlencode()

You're right, thanks! I'll edit it!

Great solution. For Python 3 compatibility, you might want to update it to use kwargs.items() instead of kwargs.iteritems()

the " ? " is getting encode when I use this code. It shows like %3Fprecio=0 (precio being the variable i'm passing)

@cursologo-gh, try using the safe parameter of urlencode.
From django docs: Use safe to pass characters which don’t require encoding. For example:

q = QueryDict(mutable=True)
q['next'] = '/a&b/'
q.urlencode(safe='/')
'next=/a%26b/'
https://docs.djangoproject.com/en/2.1/ref/request-response/#django.http.QueryDict.urlencode

Great! Tanks! (line 19: for Python 3 rename iteritems() to items())

If you are working with djangos pagination and your template takes care of the page keyword, simply adding

query.pop('page', None)

before returning the query does the trick.