Rendering dynamic variable in a django template

Question

I am trying to render dynamic Url in Django template as follows <a href={{'jibambe_site_detail/'|add: site.id}}>{{ site }}</a>. This however is returning a TemplateSyntaxError at /jibambe_sites/ add requires 2 arguments, 1 provided. What am I missing or how should I render this dynamic URL, I want it to produce something like jibambe_site_detail/1

You should really use the url tag, but have you tried href="/jibambe_site_detail/{{ site.id }}"? — user2390182
– user2390182, Commented Jan 15, 2018 at 2:38
@schwobaseggl Yes I have. The result is just jibambe_site_detail/ — E_K
– E_K, Commented Jan 15, 2018 at 2:41
That would mean that either site is None or not in the context or does not have the attribute id. — user2390182
– user2390182, Commented Jan 15, 2018 at 2:43
@schwobaseggl from your comment that I should use url tag, I have been able to resolve the issue as follows <a href={% url 'site_details' pk=site.id %}>{{ site.name }}</a> — E_K
– E_K, Commented Jan 15, 2018 at 2:56

Anupam · Accepted Answer · 2018-05-04 09:34:19Z

1

OP's strategy changed on this one but OP's original error seems to be because of the space after colon.

Use add:site.id instead of add: site.id

See a similar question about default here

answered May 4, 2018 at 9:34

Anupam

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E_K · Accepted Answer · 2018-01-15 03:03:57Z

0

From @schwobaseggl comment, I was able to add a dynamic url variable as follows <a href={% url 'site_details' pk=site.id %}>{{ site.name }}</a> then in the urls.py urlpatterns, I gave the path to jibambe_site_details a name path('jibambe_site_detail/<slug:pk>', JibambeSitesDetails.as_view(), name='site_details'),

answered Jan 15, 2018 at 3:03

E_K

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