Question

我在项目admin/testing/上这样添加了新路径urls.py

from django.views.generic.base import TemplateView
...
path('admin/', admin.site.urls),
path('admin/testing/', TemplateView.as_view(template_name='admin/testing/testing.html')),
...

，以及testing.html上的my_app/templates/admin/testing/testing.html，但我无需先登录即可访问admin/testing/。在testing.html上我有一个表单，只有当我提交表单时，它才会提示您输入我的用户名和密码。

<!-- testing.html -->
{% extends "admin/base_site.html" %}
{% load i18n static %}

{% block extrastyle %}{{ block.super }}<link rel="stylesheet" type="text/css" href="{% static "admin/css/dashboard.css" %}" />{% endblock %}

{% block coltype %}colMS{% endblock %}

{% block bodyclass %}{{ block.super }} dashboard{% endblock %}

{% block breadcrumbs %}{% endblock %}

{% block content %}
    <div id="content-main">
        <div id="testing-root"></div>

        <script crossorigin src="https://unpkg.com/react@16/umd/react.production.min.js"></script>
        <script crossorigin src="https://unpkg.com/react-dom@16/umd/react-dom.production.min.js"></script>
        <script src="https://cdnjs.cloudflare.com/ajax/libs/babel-standalone/6.26.0/babel.min.js"></script>
        <script src="https://unpkg.com/axios/dist/axios.min.js"></script>

        <script type="text/babel">
            class AdminTesting extends React.Component {
                ...
                render() {
                    const { state } = this;
                    return (
                        <div>
                            <h2>Transform Users Play Count into Ratings</h2>
                            <form>
                                Weight:<br/>
                                <input
                                    type="number"
                                    step={0.1}
                                    name="transformWeight"
                                    value={state.transformWeight}
                                    onChange={this.handleChangeFraction}
                                /><br/>
                                <input style={styles.btn} type="submit" value="Submit" onClick={this.handleRatingTransformation} />
                                { state.transformLoading && 'Please wait ...' }
                            </form>
                            ...
                        </div>
                    )
                }
            }
            ...
            ReactDOM.render(
                <AdminTesting/>,
                document.getElementById('testing-root'),
            );
        </script>
    </div>
{% endblock %}

编辑：

class TemplateView(TemplateResponseMixin, ContextMixin, View):
    """
    Render a template. Pass keyword arguments from the URLconf to the context.
    """
    def get(self, request, *args, **kwargs):
        context = self.get_context_data(**kwargs)
        return self.render_to_response(context)

Answer 1

您可以使用装饰器来限制访问权限

staff_member_required是满足您需求的好装饰器。请参考this了解更多信息

from django.contrib.admin.views.decorators import staff_member_required

path('admin/testing/', staff_member_required(TemplateView.as_view(template_name='admin/testing/testing.html')))

Answer 2

在您的urls.py中，您使用的是{strong> TemplateView 中的django.views.generic.base.TemplateView。但是它应该来自您相应的views.py，即您定义的位置。

除此之外，请使用login_required decorator进行身份验证，

from django.contrib.auth.decorators import login_required
from django.views.generic.base import TemplateView # remove this import
from .views import TemplateView # import "TemplateView" from your "views.py"

urlpatterns = [
    path('admin/', admin.site.urls),
    path('admin/testing/', login_required(
        TemplateView.as_view(template_name='admin/testing/testing.html'), login_url='/admin/')),
]

Answer 3

在您的TemplateView中尝试以下代码：

class TemplateView(TemplateResponseMixin, ContextMixin, View):
    """
    Render a template. Pass keyword arguments from the URLconf to the context.
    """
    def get(self, request, *args, **kwargs):
        if not request.user.is_authenticated():
            return render(request, 'target.html')
            #target is where you want the unauthorized logger
        else:
            context = self.get_context_data(**kwargs)
            return self.render_to_response(context)

如何在扩展的Django Admin页面上重定向未经身份验证的访问者？

3 个答案: