Python 2.7 Django 1.8中的视图

我很难将视图链接到我正在构建的电子商务模板中的基本HttpResponse

我得到的错误是404，但在看了这里和医生说的内容后，我有点困惑我错过了什么

模型

   from django.db import models

class Category(models.Model):
    parent = models.ForeignKey('self', null=True, blank=True)
    name = models.CharField(max_length=255)
    slug = models.SlugField(max_length=150)
    description = models.TextField()
    pub_date = models.DateTimeField(auto_now_add=True)
    mod_date = models.DateTimeField(auto_now=True)

class Manufacturer(models.Model):
    name = models.CharField(max_length=150)
    slug = models.SlugField(max_length=150)


class Product(models.Model):
    category = models.ForeignKey(Category)
    manufacturer = models.ForeignKey(Manufacturer)
    name = models.CharField(max_length=300)
    slug = models.SlugField(max_length=150)
    description = models.TextField()
    photo = models.ImageField(upload_to='itemphotos')
    price_in_sterling = models.DecimalField(max_digits =6, decimal_places=2)
    available = models.BooleanField()
    instock = models.IntegerField()
    pub_date = models.DateTimeField(auto_now_add=True)
    mod_date = models.DateTimeField(auto_now=True)

我的观点。py

    from django.shortcuts import render
    from django.http import HttpResponse

    def index(request):
        return HttpResponse("Hi, your view worked")

    def catagory(request):
        return HttpResponse("Hi, you are looking at the catagory landing page")

    def manufacturer(request):
        return HttpResponse("Hi, here you can shop by brand")

    def product(request):
        return HttpResponse("Hi, here you can search by product")

我的URL.py

from django.conf.urls import include, url
from django.contrib import admin


urlpatterns = [
    url(r'^productcatalog/', include('productcatalog.urls')),
    url(r'^admin/', include(admin.site.urls)),
]

应用程序产品目录中的My urls.py

from django.conf.urls import url
from . import views


urlpatterns = [
    url(r'^$', views.index, name='index'),
    url(r'^productcatalog/$', views.catagory, name='catagory'),
    url(r'^productcatalog/$', views.manufacturer, name='brand'),
    url(r'^productcatalog/$', views.product, name='products'),
]

---

我确信错误在第二个url.py文件中，但似乎无法捕捉到它。

我有另一种方法来解决这个问题

url.py

从视图导入产品
url模式=[url（r'^product/$，product）]

views.py

def产品（请求）：
view=“”某物
返回HttpResponse（视图）

---

我不记得我是否使用了render方法，但是当您在html页面中放置变量时，它确实非常有用，因为所有视图函数都使用相同的url。试着改变它

urlpatterns = [
url(r'^$', views.index, name='index'),
url(r'^productcatalog-1/$', views.catagory, name='catagory'),
url(r'^productcatalog-2/$', views.manufacturer, name='brand'),
url(r'^productcatalog-3/$', views.product, name='products'),

---

]

我试过这个，路易斯，我也犯了同样的错误。我试着把3门课从一个模型转到4页。一页链接到三个单独的页面。我也会添加模型，也许会有帮助