python - ListView 中的 Django 表单，表单 POST 时出错

Question

我正在尝试为我的网站实现聊天功能。为了做到这一点，我遵循了以下教程:https://channels.readthedocs.io/en/latest/tutorial/

然后我稍微更改了代码以实现它。到这里为止，一切都很好。现在我想将表单数据存储在数据库中，这就是问题出现的地方。但首先是我的代码:

urls.py:

    from django.urls import path
    from .views import ChatOverView

    urlpatterns = [
        path('<int:pk>/', ChatOverView.as_view(), name='chat-explicit'),
        path('', ChatOverView.as_view(), name='chat-home'),
    ]

views.py (这里有很多代码可能不需要这个问题，但由于我不知道我可以忽略其中的哪一部分，我只是发布整个文件内容):

from django.views.generic import ListView
from django.views.generic.edit import FormMixin, FormView
from django.db.models import Q
from django.urls import resolve
from django.contrib.auth.models import User
from django.shortcuts import get_object_or_404

from .models import Message
from .forms import MessageRegisterForm


class ChatOverView(ListView, FormMixin):
    model = Message
    template_name = 'chat/home-chat.html'
    form_class = MessageRegisterForm
    success_url = '/thanks/'

    def form_valid(self, form):
        form = self.get_form()
        if form.is_valid():
            data = form.cleaned_data
            return super().form_valid(form)

    def get_context_data(self, *args, **kwargs):
        context = super(ChatOverView, self).get_context_data(*args, **kwargs)

        messages_raw = reversed(Message.objects.filter(Q(sender=self.request.user) | Q(receiver=self.request.user)))
        messages = {}

        for mes in messages_raw:
            # i am receiver
            if mes.sender != self.request.user:
                if mes.sender in messages:
                    messages[mes.sender].append({"received": mes})
                else:
                    messages.update({mes.sender: [{"received": mes}]})
            # i sent
            else:
                if mes.receiver in messages:
                    messages[mes.receiver].append({"sent": mes})
                else:
                    messages.update({mes.receiver: [{"sent": mes}]})

        active_user = self.get_active_chat(messages)
        chatroom_name = self.get_chatroom_name(active_user)
        context.update(messages_data=messages, active=active_user, roomname=chatroom_name)
        return context

    def get_chatroom_name(self, active_chat):
        # my convention
        ids = [active_chat.id, self.request.user.id]
        ids.sort()
        return str(ids[0]) + '_' + str(ids[1])

    def get_active_chat(self, messages):
        url_name = resolve(self.request.path_info).url_name
        if url_name == "chat-home":
            return list(messages.keys())[0]
        else:
            pk_user = self.request.build_absolute_uri().split("/")[-2]
            user = get_object_or_404(User, pk=pk_user)
            return user

home-chat.html:

{% extends "solawi/base.html" %}
{% load define_dictfilters %}
{% block content %}
            <form class="bg-light" method="post">
                <div class="input-group">
                    {% csrf_token %}
                    {{ form }}
                    <div class="input-group-append">
                        <button type="submit" id="chat-message-submit" value="enter">send</button>
                    </div>
                </div>
            </form>    
{% endblock content %}

models.py

from django.db import models
from django.utils import timezone
from django.contrib.auth.models import User


class Message(models.Model):
    sender = models.ForeignKey(User, related_name="sender", on_delete=models.CASCADE)
    receiver = models.ForeignKey(User, related_name="receiver", on_delete=models.CASCADE)
    content = models.TextField()
    date = models.DateTimeField(default=timezone.now)

    def __str__(self):
        return self.sender.username + ' to ' + self.receiver.username

其余的文件/设置条目与上面链接的教程中的相同。

现在问题来了:提交表单时，出现以下错误:不允许的方法 (POST):/chat/
不允许的方法:/chat/
HTTP POST/chat/405 [0.00, 127.0.0.1:54424]

我该如何修复它？感谢您的帮助!!

Answer 1

ListView 实现了 get() 方法，但没有实现 post() 方法。您需要实现 post() 方法，以便 View 允许 POST 请求。

您可以子类化django.views.generic.ProcessFormView来获取它，或者，如果您确实需要ListView，那么您可以添加post( ) 方法到处理表单验证以及您需要执行的任何其他操作的类。以下是 ProcessFormView 的实现方式:

    def post(self, request, *args, **kwargs):
        """
        Handle POST requests: instantiate a form instance with the passed
        POST variables and then check if it's valid.
        """
        form = self.get_form()
        if form.is_valid():
            return self.form_valid(form)
        else:
            return self.form_invalid(form)