Django基于类的视图：如何传递其他参数到as_view方法？

问题描述：

我有一个自定义的基于类的视图

I have a custom class-based view

# myapp/views.py
from django.views.generic import *

class MyView(DetailView):
    template_name = 'detail.html'
    model = MyModel

    def get_object(self, queryset=None):
        return queryset.get(slug=self.slug)

我想传递slug参数（或视图中的其他参数）

I want to pass in the slug parameter (or other parameters to the view) like this

MyView.as_view(slug='hello_world')

我需要覆盖任何方法才能实现吗？

Do I need to override any methods to be able to do this?

答

传递给 as_view 方法的每个参数都是View类的一个实例变量。这意味着添加 slug 作为参数，您必须将其创建为子类中的实例变量：

Every parameter that's passed to the as_view method is an instance variable of the View class. That means to add slug as a parameter you have to create it as an instance variable in your sub-class:

# myapp/views.py
from django.views.generic import *

class MyView(DetailView):
    template_name = 'detail.html'
    model = MyModel
    # additional parameters
    slug = None

    def get_object(self, queryset=None):
        return queryset.get(slug=self.slug)

应该使 MyView.as_view（slug ='hello_world'） work。

如果您通过关键字传递变量，请使用Erikkson先生的建议： https://stackoverflow.com/a/11494666/9903

If you're passing the variables through keywords, use what Mr Erikkson suggested: https://stackoverflow.com/a/11494666/9903

Django基于类的视图：如何传递其他参数到as_view方法？

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