在 Django 中使用 DetailView 查看用户配置文件

我需要简单博客网站的帮助。用户可以写帖子的地方。每个帖子作者都必须登录才能撰写帖子。但读者不必登录。

我正在使用 django DetailView 来允许读者查看帖子作者的个人资料。

个人资料视图将包含作者的详细信息以及作者撰写的其他帖子。

我使用以下逻辑来制作视图https://chriskief.com/2012/12/29/django-generic-detailview-without-a-pk-or-slug/

以下是详细信息：

所有这些都在我网站的帐户应用程序中。下面的代码被破坏并给出错误

错误消息的图像

帐户/视图.py：

    class ProfileView(DetailView):
            model = User
            template_name = 'accounts/profile.html'

            def get_user_profile(self, username):   
                return get_object_or_404(User, pk=username)
     #I know pk=username is not correct. I am not sure what to put pk=?

  # I was able to get the writers other posts using the code below. I did not have to show this code for this question. But just to show you that the pk above has to be username. Or Else the code below won't work(I guess)        
            def get_context_data(self, **kwargs):
                context = super(ProfileView, self).get_context_data(**kwargs)
                context['post_list'] = Post.objects.filter(user__username__iexact=self.kwargs.get('username'))
                return context  

我的模型如下

帐户/模型.py：

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

    class UserProfile(models.Model):
        user = models.OneToOneField(User, on_delete=models.CASCADE)
        city = models.CharField(max_length=100)
        country = models.CharField(max_length=100)
        bio = models.TextField(default='', blank=True)
        profile_image = models.ImageField(default='', blank=True, null='')


        def __str__(self):
            return "Profile of {}".format(self.user.username)

下面是我的模板

    {% extends 'base.html' %}

    {% block body %}
    <div class="content">
        <h1>{{user.first_name}} {{user.last_name}}</h1>
        <p>            
            <br/>From: City, Country
            <br/>Bio
            <br/>Profile Image
        </p>
    </div>


    {% endblock %}

帐户/网址.py

from django.conf.urls import url
from . import views
from django.contrib.auth.views import LoginView, LogoutView

app_name = 'accounts'

urlpatterns = [
    url(r'^login/$', LoginView.as_view(template_name='accounts/login.html'), name='login'),
    url(r'^logout/$', LogoutView.as_view(), name='logout'),
    url(r'^signup/$', views.SignUp.as_view(), name='signup'),
    url(r'^profile/(?P<username>[-\w]+)/$', views.ProfileView.as_view(), name='profile'),

]

错误消息文本

/accounts/profile/3/ 处的 AttributeError 必须使用对象 pk 或 slug 调用通用详细信息视图 ProfileView。请求方法：GET 请求 URL：http : //127.0.0.1 :
8000/accounts/profile/3/ Django 版本：1.11 异常类型：AttributeError 异常值：必须使用对象 pk 或 slug 调用通用详细信息视图 ProfileView。异常位置：C:\Users\Samir\Miniconda3\envs\MyDjangoEnv\lib\site-packages\django-1.11-py3.6.egg\django\views\generic\detail.py in get_object, line 49 Python Executable: C :\Users\Samir\Miniconda3\envs\MyDjangoEnv\python.exe Python 版本：3.6.4 Python 路径：........