【发布时间】:2018-03-11 21:13:00
【问题描述】:
我是 django 的新手,我目前正在做一个项目。这是一个人们可以寻找房屋出租的网站。用户将能够创建帐户、搜索要出租的房屋以及创建他们想要出租的房屋的列表。
我创建了一个模型来保存有关用户想要出租的房屋的所有信息。我需要过滤这些信息并在他们的个人资料中显示每个用户的列表。我在网上搜索过,但还没有解决方案。
真的需要帮助。
models.py
from django.db import models
from django.contrib.auth.models import User
class Myhouses(models.Model):
Available = 'A'
Not_Available = 'NA'
Availability = (
(Available, 'Available'),
(Not_Available, 'Not_Available'),
)
name_of_accomodation = models.CharField(max_length=200)
type_of_room = models.CharField(max_length=200)
house_rent = models.IntegerField()
availability = models.CharField(max_length=2, choices=Availability, default=Available,)
location = models.CharField(max_length=200)
nearest_institution = models.CharField(max_length=200)
description = models.TextField(blank=True)
image = models.ImageField(upload_to='profile_image')
author = models.ForeignKey(User, on_delete=models.SET_NULL, null=True, blank=True, related_name='author')
def __str__(self):
return self.name_of_accomodation
view.py
class ListingByUser(LoginRequiredMixin, generic.ListView):
model = Myhouses
template_name ='houses/ListingByUser.html'
paginate_by = 10
def get_queryset(self):
return Myhouses.objects.filter(author=self.request.user)
urls.py
from django.conf.urls import url, include
from . import views
from django.contrib.auth.models import User
urlpatterns = [
url(r'^addlisting/$', views.addlisting, name='addlisting'),
url(r'^mylisting/', views.ListingByUser.as_view(), name='ListingByUser')
]
模板
<ul>
{% for houses in myhouses_list %}
<li>{{ houses.name_of_accomodation }}</li>
{%endfor %}
</ul>
【问题讨论】:
-
没有解决办法是什么?您发布的代码有什么问题?