I'm trying to generate a link where the URL of the current page gets passed in the querystring of the link. Like so: <a href="http:// www.example.com?variable=http%3A%2F%2Fwww.example2.com%2Farg1%2Farg2%2F"></a>.
I would have thought that I could do this relatively easily with template tags, so I've tried this: <% url arg1, arg2 as the_url %> <a href="http://www.example.com?variable={{ the_url|urlencode }}"></a> However, for whatever reason, the urlencode filter isn't applied when the_url renders. Is there a way to do this directly in the template, or do I have to generate the encoded URL in the view and pass it to the template? -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-us...@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
