Question

我有一个电影和电视节目的数据库。我想像这样过滤这些： / productions / = index（all），/ productions / films / = only films，and productions / series / = only tv shows

## 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')

这样的事情最好的做法是什么？在views.py中为每个方法创建一个新方法，或者我可以重用main方法，并通过某种方式解析URL段来调用if（productions.is_movie）之类的东西吗？

Answer 1

我会从url中捕获字符串，如下所示：

urlpatterns = patterns('',
    url(r'^(?<query>(films|series|))/$', views.IndexView.as_view(), name='films_series'),
)

然后，在get_queryset()方法中，我会检查你是否需要归还所有，电影或连续剧：

class IndexView(generic.ListView):
    template_name = 'productions/index.html'
    context_object_name = 'productions_list'

    def get_queryset(self):
        # analyze `self.kwargs` and decide should you filter or not, just for example:
        is_all = self.kwargs['query'] == ''
        is_movie = self.kwargs['query'] == 'films' 
        is_series = self.kwargs['query'] == 'series'

        return Production.objects.order_by('-release')  # TODO: filter movies or series

按网址段过滤内容

1 个答案: