2015-07-20 49 views
2

我無法將視圖鏈接到我正在構建的電子商務模板中的基本HttpResponse。Django中的視圖1.8

我得到的錯誤是404,但希望通過在這裏和追求的文檔說我是什麼,我已經錯過了一個有點困惑,

示範

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views.py中

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urls.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)), 
] 

從應用produtcatalog

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urls.py我相信錯誤是在第二urls.py文件,但似乎無法抓住它。

回答

0

以及我對這個

urls.py另一種方法 from views import product urlpatterns=[url(r'^product/$', product)]

views.py def product(request): view="""<html><head></head><body><p>something</p></body></html> """ return HttpResponse(view)

我不記得,如果我用了渲染方法,但真正有用的特別是當你把變量放在html頁面裏

+0

我想這路易斯和我拿出同樣的錯誤。我試圖從一個模型傳遞3個類到4個頁面。一頁鏈接到三個單獨的頁面。我也將添加模型也許這將有所幫助 – AaronC

0

你對所有的視圖函數都使用相同的url。嘗試改變它。

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'), 

]