url - django cannot get request from hyperlink -

url - django cannot get request from hyperlink -

i have made hyperlink this

<a href="{% url "csv_download" %}?format="csv"">download csv</a>

and url is

url(r'^download/(?p<format>)',singlenewsview.as_view(), name="csv_download")

in view have method format , if there format=csv want query. dont know how format. when run above script gives get() takes 3 arguments (2 given). can help??

your url pattern looking named parameter, you're passing querystring parameter.

if want pattern be:

/download/csv/

then you'll need alter phone call url to:

<a href="{% url "csv_download" "csv" %}">download csv</a>

if want value querystring instead, you'l need alter url pattern to:

url(r'^download/$',singlenewsview.as_view(), name="csv_download")

and can pick "format" parameter using:

format = request.get.get('format')

in either case, you're missing pattern terminator, "/$", should be:

url(r'^download/(?p<format>)/$',singlenewsview.as_view(), name="csv_download")

django url get