django의 동적 URL

Question

저는 장고를 처음 사용했습니다. 제발 빠른 도움을 많이 부탁드립니다.

url(r'^shops/(?P<Newshop_id>[0-9]+)/$', views.shop_single, name='singleshop') 

이 경로는 저에게 페이지가 없습니다. 오류가 발생했습니다.

models.py

from __future__ import unicode_literals 
from django.db import models 

class SliderTitle(models.Model): 
    slider_title = models.CharField(max_length=20) 
    def __str__(self): 
     return self.slider_title 

class Slider(models.Model): 
    slider_type = models.OneToOneField(SliderTitle) 
    slider = models.FileField(blank=True) 
    def __str__(self): 
     return str(self.slider_type) 

class ShopCategories(models.Model): 
    category = models.CharField(max_length=50, unique=True) 
    def __str__(self): 
     return self.category 

class NewShop(models.Model): 
    category = models.ForeignKey(ShopCategories) 
    main_image = models.FileField() 
    name = models.CharField(max_length=100, unique=True) 
    tagline = models.CharField(max_length=50, default='Enter tagline here2') 
    description = models.TextField(default='enter shop description') 
    shop_image = models.FileField() 
    def __str__(self): 
     return self.name 

urls.py

urlpatterns = [ 
    url(r'^admin/', admin.site.urls), 
    url(r'^$', views.homepage, name='homepage'), 
    url(r'^about/', views.about, name='about'), 
    url(r'^shops/(?P<Newshop_id>[0-9]+)/$', views.shop_single, name='singleshop') 

views.py

def shop_single(request, Newshop_id): 
    cat1 = NewShop.objects.filter(category_id=1) 
    cat2 = NewShop.objects.filter(category_id=2) 
    cat3 = NewShop.objects.filter(category_id=3) 
    cat4 = NewShop.objects.filter(category_id=4) 
    name1 = ShopCategories.objects.filter(id=1) 
    name2 = ShopCategories.objects.filter(id=2) 
    name3 = ShopCategories.objects.filter(id=3) 
    name4 = ShopCategories.objects.filter(id=4) 

    return render_to_response('shop_single.html', {'shop_name1': name1, 'shop_name2': name2, 'shop_name3': name3, 
              'shop_name4': name4, 'Shop_cat1': cat1, 'Shop_cat2': cat2, 
              'Shop_cat3': cat3, 
              'Shop_cat4': cat4, }) 

my DB from phpmyadin

Answer 1

당신이 결코보기에 newshop_id 사용되지 않습니다 생각합니다.

그래서 당신의 전망과 당신의 URL에 그 newshop_id를 전달할 필요가 없습니다.