title: Altering one query parameter in a url (Django)
body: <p>I have a search page that takes a variety of parameters. I want to create a new URL by just altering one parameter in the query. Is there an easy way to do this - something like:</p>
  
  <pre><code># example request url
  http://example.com/search?q=foo&amp;option=bar&amp;option2=baz&amp;change=before
  
  # ideal template code
  {% url_with change 'after' %}
  
  # resulting url
  http://example.com/search?q=foo&amp;option=bar&amp;option2=baz&amp;change=after
  </code></pre>
  
  <p>So this would take the request url, alter one query parameter and then return the new url. Similar to what can be achieved in Perl's Catalyst using <code>$c-&gt;uri_with({change =&gt; 'after'})</code>.</p>
  
  <p>Or is there a better way?</p>
  
  <p>[UPDATED: removed references to pagination]</p>