Passing a variable from template in Django

Question

I want to create a generic country view that populates data based on the what country i clicked on a map. The current URL looks like:

# Country View URL 
   url(r'^country/' , 'wiki.views.country',
        name = 'wiki_country'),

And the view is:

def country(request):
    return render_to_response("wiki/country.html")

This is fine if I want a separate page for each country. I was reading elsewhere that there is no simple way to get a variable from a template in Django. What I would want is for the link in the template not only have the URL, but also a value ("Country Name") that then allows me to dynamically populate a single countryView template.

Aamir Rind · Accepted Answer · 2014-06-25 17:52:37Z

1

For specific country which will take id and name of country you can define the URL pattern as:

url(r'^country/(?P<id>\d+)/(?P<name>\w+)' , 'wiki.views.country', name='wiki_country'),

The view will be:

def country(request, id, name):
    country = Country.objects.get(id=id, name=name)
    return render_to_response("wiki/country.html", {'country': country})

In listing template you can have a link as:

{% for country in countries %}
    <a href="{% url 'wiki_country' country.id country.name %}">{{ country.name }}</a>
{% endfor %}

Hope this will lead you somewhere.

answered Jun 25, 2014 at 17:52

Aamir Rind

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2 Comments

bamabacho Over a year ago

BTW, do I need {{country.name}} in the listing template? Since this is the place the user will select the country, ought I not to just hard code it?

bamabacho Over a year ago

Ok, I think this will work. Still messing about with the model, but my question is where does the name and id of the country go so that the instance can be made based on those values? In the block on the template surely, correct?

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