How to add the current query string to an URL in a Django template?

user_612606 0
Viewed 39360

When I load a page, there is a link "sameLink" that I want to append to it the query string of its containing page.

I have following URL: 

somedomain/reporting/article-by-month?variable1=2008

How can I do that?

django python
7 Answers

Informed by other answers but not needing the request passed in and only updates existing parameters.

@register.simple_tag(takes_context=True)
def querystring(context, **kwargs):
    """
    Creates a URL (containing only the querystring [including "?"]) derived
    from the current URL's querystring, by updating it with the provided
    keyword arguments.

    Example (imagine URL is ``/abc/?gender=male&name=Tim``)::

        {% querystring "name"="Diego" "age"=20 %}
        ?name=Diego&gender=male&age=20
    """
    request = context['request']
    updated = request.GET.copy()
    for k, v in kwargs.items():  # have to iterate over and not use .update as it's a QueryDict not a dict
        updated[k] = v

    return '?{}'.format(updated.urlencode()) if updated else ''
user_2381660

Just a workaround if you don't want loops or string concatenations.

String representation of url is something like '<WSGIRequest: GET \'our_url\'>' So if you want our_url, you just need to use regex for it.

our_url = re.search(r'\/.*(?=\'>)', str(request)).group()
user_43914010
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