django templates : how to expand a variable into the string argument for the built-in tag `url`

Question

I am trying to make the following template snippet work:

<ul>
  {% for name,label in entries.items %}
    <li><a href="{% url name %}">{{ label }}</a></li>
  {% endfor %}
</ul>

As you can see name is a variable that I need to expand into string before passing it to the url built-in tag.

Unfortunately the aforementioned template snippet results in the following exception:

Exception Type: TemplateSyntaxError
Exception Value:    
Caught NoReverseMatch while rendering: Reverse for 'name' with arguments '()' and keyword arguments '{}' not found.

Any ideas or alternative methods on how to achieve this?

bruno desthuilliers · Accepted Answer · 2013-12-16 16:34:46Z

3

Your code should work for Django >= 1.4 (iirc) - assuming name resolves to a valid url name and this url don't need args or kwargs. For Django 1.3x you have to add {% load url from future %} to your template for this to work. For django < 1.3 you're out of luck (well, you'll have to write a custom tag) - but you should possibly consider porting your code anyway.

answered Dec 16, 2013 at 16:34

bruno desthuilliers

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1 Comment

fstab Over a year ago

This is the answer I was looking for. It works! Thanks! :) (I am using django 1.3.7)

niekas · Accepted Answer · 2013-12-16 16:37:08Z

3

You can pass variable to {% url %} templatetag. If name is valid urlpattern name this snippet should work.

If you need to convert name to a valid urlpattern name, you can do it by using custom filter, like this:

{% url name|my_filter %}

edited Dec 16, 2013 at 16:37

answered Dec 16, 2013 at 16:17

niekas

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2 Comments

fstab Over a year ago

Thanks for the answer. It seems to be a bit redundant the creation of a custom filter that just calls the django url resolver function; this is why I was hoping for a solution within the template.

bruno desthuilliers Over a year ago

@niekas: the exemple syntax you posted is for filters, not tags.

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