django template url - append variable to string

Question

Im trying to add a variable to the end of a string and am getting issues with reverse match, from what I gather the below should work

{% for type in type_links %}
<li><a href="{% url 'sites:site_list' 'all_'|add:type.id %}">{{ type.subnet_type }}</a></li>
{% endfor %}

I should have a url that has "all_1" for example in it.

however I am getting a reverse url error

Reverse for 'site_list' with arguments '('',)' not found. 1 pattern(s) tried: ['sites/site_list/(?P<display_filter>[\\w\\-]+)$']

is this correct way to add a variable to end the of the string?

EDIT: url pattern, the pattern works as I have tested it manually before trying to create the URLs dyanmically

url(r'^site_list/(?P<display_filter>[\w\-]+)$', views.site_list, name='site_list'),

Thanks

Could not parse the remainder: '{{type.id}}' from ''all_'{{type.id}}' I believe that its because its now seeing those as two entries not one joined entry? — AlexW
– AlexW, Commented Jan 8, 2018 at 10:56

itzMEonTV · Accepted Answer · 2018-01-08 11:51:30Z

1

You can do

{% for type in type_links %}
<li>
    {% with type.id|stringformat:"s" as str_obj_id %}
      {% with 'all_'|add:str_obj_id as extra_param %}
        <a href="{% url 'sites:site_list' extra_param %}">{{ type.subnet_type }}</a>
      {% endwith %}
   {% endwith %} 
</li>
{% endfor %}

edited Jan 8, 2018 at 11:51

answered Jan 8, 2018 at 11:24

itzMEonTV

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2 Comments

AlexW Over a year ago

im still getting Reverse for 'site_list' with arguments '('',)' not found. 1 pattern(s) tried: ['sites/site_list/(?P<display_filter>[\\w\\-]+)$']

AlexW Over a year ago

when I put {{ type_links}} I get the data, and if I just loop the IDs it also works

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