Add url parameters to a url in django view

Question

I have a django view that lists several urls on external sites.

When I render them I would like to add a few url parameters to each.

These urls are to an external system and thus not listed in my urls.py. Furthermore, some of the links have a hash '#' so it is not as easy as appending a few parameters to the end of the string.

Based on these requirements it seems the url template tag will not be a good fit. I was wondering if there is a custom filter out that to do this.

Why would you need the url tag? It would help if you could create an minimal reproducible example — Sayse
– Sayse, Commented Jun 22, 2016 at 12:21
It's not very clear what you are asking. The url template tag is for generating URLs that are within your own site/project. If you want to add links to external sites then why not just put them in the HTML, URL parameters and all? — solarissmoke
– solarissmoke, Commented Jun 22, 2016 at 12:23
This seems odd to me why would you need a thirdparty application to solve this problem. Seems like typing ?foo=bar&pop=lol is a lot less key strokes then the pip install + add_query="foo=bar" etc. — marcusshep
– marcusshep, Commented Jun 22, 2016 at 20:47
@marcusshep that's not entirely true. what about url fragments(#)? the params need to come before it. what if there are already url params? you don't need to add the '?'. I agree it's not a lot of code but there are definitely edge cases here. — elewinso
– elewinso, Commented Jun 23, 2016 at 15:00

ohrstrom · Accepted Answer · 2016-06-22 12:32:04Z

1

You don't need Django's url tag here. The url tag is to resolve to URLs that belong to your application.

However, there is the nice django-spurl library. It allows you to handle query parameters via template tags.

An example from the documentation to add query parameters:

{% spurl base="http://example.com/?foo=bar" add_query="bar=baz" %}

<!--
will result in: http://example.com?foo=bar&bar=baz 
-->

answered Jun 22, 2016 at 12:32

ohrstrom

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