Django 推文应用登出功能异常问题求助

问题概述 #

开发基于Python 3.12.0 + Django的推文应用时，登录功能正常，但登出操作失效。点击登出链接（http://127.0.0.1:8000/logout/?next=/）时出现「无法访问此网站」错误，无法完成账号登出。

相关代码 #

urls.py #

from django.contrib import admin
from django.urls import path, include 

urlpatterns = [
    path('admin/', admin.site.urls),
    path("",include('tweetapp.urls')),
    path("",include('django.contrib.auth.urls')),
]

views.py #

from django.shortcuts import render, redirect
from . import models
from django.urls import reverse
from tweetapp.forms import AddTweetForm, AddTweetModelForm

def listtweet(request):
    all_tweets = models.Tweet.objects.all()
    tweet_dict = {"tweets":all_tweets}
    return render(request,'tweetapp/listtweet.html',context=tweet_dict)

def addtweet(request):
    if request.POST:
        nickname = request.POST["nickname"]
        message = request.POST["message"]
        models.Tweet.objects.create(nickname=nickname, message=message)
        return redirect(reverse('tweetapp:listtweet'))
    else:
        return render(request,'tweetapp/addtweet.html')
    
def addtweetbyform(request):
    if request.method == "POST":
        form = AddTweetForm(request.POST) 
        if form.is_valid():
            nickname = form.cleaned_data["nickname_input"]
            message = form.cleaned_data["message_input"]
            models.Tweet.objects.create(nickname=nickname, message=message
                                        )
            return redirect(reverse('tweetapp:listtweet'))
        else:
            print("error in form")
            return render(request,'tweetapp/addtweetbyform.html', context={"form":form})
    else:
        form = AddTweetForm()
        return render(request,'tweetapp/addtweetbyform.html', context={"form":form})

def addtweetbymodelform(request):
    if request.method == "POST":
        form = AddTweetModelForm(request.POST) 
        if form.is_valid():
            print(form.cleaned_data)
            return redirect(reverse('tweetapp:listtweet'))
        else:
            print("error in form")
            return render(request,'tweetapp/addtweetbymodelform.html', context={"form":form})
    else:
        form = AddTweetModelForm()
        return render(request,'tweetapp/addtweetbymodelform.html', context={"form":form})

原base.html中的登出代码 #

<li class="nav-item">
  <a class="nav-link" href="{% url 'login' %}">Login</a>
</li>
<li class="nav-item">
  <a class="nav-link" href="{% url 'logout' %}?next=/">Logout</a> 
</li>

解决方法 #

Django内置的logout视图默认仅接受POST请求（出于CSRF安全防护考虑），原代码中使用<a>标签发起的是GET请求，因此无法正常触发登出逻辑。将登出链接替换为POST表单形式，带上CSRF令牌即可解决问题：

修改后的base.html登出代码：

<form method="post" action="{% url 'logout' %}">
  {% csrf_token %}
  <button class="nav-link" type="submit">Logout</button>
</form>

内容的提问来源于stack exchange，提问作者airbone7