django url in template correct way to add parameter

Question

in views.py

class LaViewSet(viewsets.ModelViewSet):

    serializer_class = IlSerializer

    def get_queryset(self):
        ilfiltro = self.kwargs['miopar']
        return models.Pippo.objects.filter(pippo=ilfiltro)

in url.py

url(r'^pippo/(?P<miopar>.+)', views.LaViewSet.as_view({'get': 'list'}), name="Serializzata"),

this is a working url:

http://127.0.0.1:8000/pippo/1

but if I put in a template:

{% url '1' 'Serializzata' %};

or

{% url 'Serializzata'?1 %};

keep getting this error:

TemplateSyntaxError: Could not parse the remainder: '?1' from ''Serializzata'?1'

Leistungsabfall · Accepted Answer · 2015-10-21 19:18:43Z

3

From the docs:

url

Returns an absolute path reference (a URL without the domain name) matching a given view and optional parameters. Any special characters in the resulting path will be encoded using iri_to_uri().

This is a way to output links without violating the DRY principle by having to hard-code URLs in your templates:

{% url 'some-url-name' v1 v2 %}

So in your case:

{% url 'Serializzata' 1 %}

answered Oct 21, 2015 at 19:18

Leistungsabfall

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SilentDev · Accepted Answer · 2015-10-21 19:22:20Z

1

Try this:

<a href="{% url 'Serializzata' 1 %}">

edited Oct 21, 2015 at 19:22

answered Oct 21, 2015 at 19:16

SilentDev

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