Django url在模板中添加param的正确方法

Question

在视图.py

class LaViewSet(viewsets.ModelViewSet):

    serializer_class = IlSerializer

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

在网址.py

^{pr2}$

这是一个有效的url:

http://127.0.0.1:8000/pippo/1

但如果我放一个模板：

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

或者

{% url 'Serializzata'?1 %};

继续出现此错误：

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

Answer 1

从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 %}

所以在你的情况下：

{% url 'Serializzata' 1 %}

Answer 2

试试这个：

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