Django 按url段筛选内容
我有一个电影和电视节目的数据库。我想这样过滤这些: /productions/=索引(全部),/productions/films/=仅电影和/productions/series/=仅电视节目Django 按url段筛选内容,django,python-3.x,django-templates,django-views,Django,Python 3.x,Django Templates,Django Views,我有一个电影和电视节目的数据库。我想这样过滤这些: /productions/=索引(全部),/productions/films/=仅电影和/productions/series/=仅电视节目 ## urls.py from django.conf.urls import patterns, url from productions import views urlpatterns = patterns('', url(r'^$', views.IndexView.as_view(
## urls.py
from django.conf.urls import patterns, url
from productions import views
urlpatterns = patterns('',
url(r'^$', views.IndexView.as_view(), name='index'),
url(r'^films/$', views.IndexView.as_view(), name='films'),
url(r'^series/$', views.IndexView.as_view(), name='series'),
)
## views.py
from django.shortcuts import render, get_object_or_404
from django.http import HttpResponseRedirect, HttpResponse
from django.core.urlresolvers import reverse
from django.views import generic
from productions.models import Production, Director
class IndexView(generic.ListView):
template_name = 'productions/index.html'
context_object_name = 'productions_list'
def get_queryset(self):
return Production.objects.order_by('-release')
对于这样的事情,最好的做法是什么?在views.py中为每一个创建一个新方法,或者我可以重用主方法,并通过解析URL段调用类似if(productions.is\u movie)的东西吗?我会从URL捕获字符串,如下所示:
urlpatterns = patterns('',
url(r'^(?<query>(films|series|))/$', views.IndexView.as_view(), name='films_series'),
)
我将从url捕获字符串,如下所示:
urlpatterns = patterns('',
url(r'^(?<query>(films|series|))/$', views.IndexView.as_view(), name='films_series'),
)