Question

如何从中间件文件的reason模型中将UserBanning传递到banned.html？几乎所有东西都可以正常工作，但是我似乎无法从模型中获得reason并无法在模板banned.html中显示，而且不确定方式，因此只要深入学习中间件，任何帮助都将是惊人的。我应该改用process_request()吗？

谢谢

models.py：

from django.db import models
from django.contrib.auth.models import User
from django.conf import settings

class UserBanning(models.Model):
    user = models.ForeignKey(User, verbose_name="Username", help_text="Choose Username", on_delete=models.CASCADE)
    ban = models.BooleanField(default=True, verbose_name="Ban", help_text="Users Bans")
    reason = models.CharField(max_length=500, blank=True)

    class Meta:
        verbose_name_plural = "User Banning"
        ordering = ('user',)

    def __str__(self):
        return f"{self.user}"

middleware.py：

from .models import UserBanning
from django.shortcuts import render


class BanManagement():
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        banned = UserBanning.objects.all()
        context = {
            'banned': banned,
        }

        if(banned.filter(ban=True, user_id=request.user.id)):
            return render(request, "account/banned.html", context)
        else:
        response = self.get_response(request)
        return response

banned.html：

{% extends "base.html" %}

{% block content %}
<p>Your account has been banned. Reason: {{ banned.reason }}</p>
{% endblock content %}

Answer 1

您快完成了。剩下的问题是，您需要在设置filter之前在request.user.id上context，以便上下文仅包含此特定用户的禁令。像这样：

    banned = UserBanning.objects.filter(ban=True, user=request.user)

    if banned:
        context = {'banned': banned[0]}
        return render(request, "account/banned.html", context)
    else:

我的Django中间件无法正常工作

1 个答案: