为什么要在Django中使用基本视图？

PythonAnywhere

为什么在这种情况下，为什么要在Django中使用基本视图

from django.http import HttpResponse
from django.views import View

class MyView(View):

    def get(self, request, *args, **kwargs):
        return HttpResponse('Hello, World!')

可以写成

def get(request):
    return HttpResponse('Hello, World!')

基本视图与功能视图的优点是什么？

2ps

种种原因。

您想使用专门的视图，例如TemplateView@pythonista提到的那样，它使编写视图变得更加容易。例如，
```
class MyTemplateView(TemplateView):
    template_name = 'template.html'
```

当您有类似的行为时，您希望有所隔离。例如，您希望基于类的View同时处理表单呈现和表单发布：

class MyFormView(TemplateView):
    template_name = 'form.html'

    def get(self, request, *args, **kwargs):
        return super(MyFormView, self).get(request, *args, **kwargs)

    def post(self, request, *args, **kwargs:
        value1 = request.POST.get('value1')
        value2 = request.POST.get('value2')
        # handle the post values
        return super(MyFormView, self).get(request, *args, **kwargs)

您拥有REST端点，并且希望在一个基于类的视图中隔离GET / POST / PUT / DELETE的所有代码。

    class RestEndpoint(View):
        def __init__(self):
            super(RestEndpoint, self).__init__()
            self.model = MyModel

        def get(request, n_id, *args, **kwargs):
            x = self.model.objects.get(id=n_id)
            return JsonResponse(x.to_json())

        def put(self, request, *args, **kwargs):
            data = json.loads(request.body)
            x = self.model(**data)
            x.save()
            return JsonResponse(x.to_json())

        def post(self, request, n_id, *args, **kwargs):
            data = json.loads(request.body)
            x = self.model.objects.get(id=n_id)
            for key, value in data.items():
                setattr(x, key, value)
            x.save()
            return JsonResponse(x.to_json())

        def delete(self, request, n_id, *args, **kwargs):
            self.model.objects.filter(id=n_id).delete()
            return JsonResponse({})