我正在寻找使用Django构建一个小型的“Twitter风格”网站来处理事情,并决定尝试允许多个用户编辑每个帖子(最终基于权限)。现在我正在努力的是访问每个用户的帖子。下面是我的模型,视图和模板的代码,显示所有用户“此处没有帖子”。我希望能够显示用户所拥有的所有帖子,而且似乎无处可去:
models.py
from django.db import models
class User(models.Model):
username = models.CharField(max_length = 200)
email = models.EmailField(max_length = 75)
password = models.CharField(max_length = 64)
created_date = models.DateTimeField('date created')
def __unicode__(self):
return self.username
class Meta:
ordering = ('created_date',)
class Post(models.Model):
users = models.ManyToManyField(User)
title = models.CharField(max_length = 300)
post = models.TextField()
posted_date = models.DateTimeField('date created')
votes = models.IntegerField()
def __unicode__(self):
return self.title
class Meta:
ordering = ('posted_date',)
views.py
from django.shortcuts import render, get_object_or_404
from django.http import HttpResponse
from users.models import User, Post
def index(request):
latest_user_list = User.objects.order_by('username')[:5]
context = {'latest_user_list': latest_user_list}
return render(request, 'users/index.html', context)
def detail(request, user_id):
user = get_object_or_404(User, pk=user_id)
post_list = Post.objects.filter(id == user.id)
return render(request, 'users/detail.html', {'user': user, 'post': post_list})
urls.py
from django.conf.urls import patterns, url
from users import views
urlpatterns = patterns('',
url(r'^$', views.index, name='index'),
url(r'^(?P<user_id>\d+)/$', views.detail, name='detail'),
)
(模板) - detail.html
<h1>{{ user.username }}</h1>
{% if post_list %}
<ul>
{% for post in post_list%}
<li>{{ post.title }}</li>
{% endfor %}
</ul>
{% else %}
<p> There aint no posts here </p>
{% endif %}
答案 0 :(得分:2)
您传递给模板的变量称为post
而不是post_list
。
答案 1 :(得分:2)
在视图中更改列表对象的名称。
def detail(request, user_id):
user = get_object_or_404(User, pk=user_id)
post_list = Post.objects.filter(id == user.id)
return render(request, 'users/detail.html', {'user': user, 'post_list': post_list})