Django 动态扩展基于 class 的视图

Question

在 Django 的基于 class 的视图中，是否有动态选择我想扩展哪个视图的好方法？

现在我有 2 个视图（更新和视图）如下，我想将其合并为 1 个：

from django.views.generic import UpdateView, DetailView
from guardian.mixins import PermissionRequiredMixin
 
class MyUpdateView(PermissionRequiredMixin, UpdateView):
    model = MyModel
    permission_required = 'MyApp.change_mymodel'
    template_name = "MyApp/update.html"
    fields = ["name", "Type", "description"]

class MyDetailView(PermissionRequiredMixin, DetailView):
    model = MyModel
    permission_required = 'MyApp.view_mymodel'
    template_name = "MyApp/view.html"

我想做的是有1个视图，1个url，当访问视图时如果用户有更改权限，应该使用UpdateView，如果他们有查看权限使用DetailView，如果它们都没有显示某种错误。

有没有办法让它起作用？

编辑：最终解决方案：

class CombinedView(PermissionRequiredMixin, DetailView, UpdateView):
model = MyModel
fields = ["name", "Type", "description"]

def get_required_permissions(self, request=None):
    if (self.request.user).has_perm('MyApp.change_mymodel', self.get_object()):
        return ['MyApp.change_mymodel']
    return ['MyApp.view_mymodel']

def get_template_names(self):
    if (self.request.user).has_perm('MyApp.change_mymodel', self.get_object()):
        return ["MyApp/update.html"]
    return ["MyApp/view.html"]

Answer 1

重写 get_template_names() 方法和 get_required_permissions() 方法，

class CombinedView(PermissionRequiredMixin, DetailView, UpdateView):
    model = MyModel
    fields = ["name", "Type", "description"]

    <b>def get_required_permissions(self, request=None):  # alternate to `permission_required`
        if self.request.method == 'POST':
            return ['MyApp.change_mymodel']
        return ['MyApp.view_mymodel']

    def get_template_names(self):  # alternate to `template_name`
        if self.request.method == 'POST':
            return ["MyApp/update.html"]
        return ["MyApp/view.html"]</b>

Django 动态扩展基于 class 的视图

Django Extending class based view dynamically

django

django-guardian