在Django 1.7中动态过滤ListView CBV

Question

我已经阅读了关于动态过滤 ListView 的官方文档,但我仍然对如何实际使用它感到困惑.

我目前有一个简单的模型,我们称之为Scholarship:

class Scholarship(models.Model):
    title = models.CharField(max_length=255)
    submitted_date = models.DateField(auto_now=True, verbose_name='Date Submitted')
    EXPERIENCE_LEVEL_CHOICES = (
        ('A', 'Any'),
        ('S', 'Student'),
        ('G', 'Graduate')
    )
    experience_level = models.CharField(max_length=1, choices=EXPERIENCE_LEVEL_CHOICES, default='A')

我有一个页面,我正在使用ListView显示所有这些奖学金:

views.py

from django.views.generic import ListView
from .models import Scholarship


class ScholarshipDirectoryView(ListView):
    model = Scholarship
    template_name = 'scholarship-directory.html'

urls.py

from django.conf.urls import patterns, url

from .views import ScholarshipDirectoryView

urlpatterns = patterns('',
    url(r'^$', ScholarshipDirectoryView.as_view(), name='scholarship_directory'),
)

我正在尝试在网站的主页上生成链接,该链接将返回此ListView的过滤版本.例如,如果有人点击"为研究生提供奖学金"链接,则只会显示奖学金experience_level='G'.

通过shell - >返回此查询集没有问题 Scholarship.objects.filter(experience_level__exact='G')

我只是不确定如何通过下拉列表或URL动态过滤ListView.不想使用插件,而是了解动态查询/过滤在Django中的工作原理.

Answer 1

首先,您需要更改urls.py,以便将经验作为参数传递.像这样的东西:

urlpatterns = patterns('',
    url(r'^(?P<exp>[ASG])$', ScholarshipDirectoryView.as_view(), name='scholarship_directory'),
)

(如果/ A或/ S或/ G未通过,以上将返回404)

现在,在 kwargsCBV的属性中,我们将有一个命名的kwarg exp,该get_queryset方法可以根据经验级别进行过滤.

class ScholarshipDirectoryView(ListView):
    model = Scholarship
    template_name = 'scholarship-directory.html'

    def get_queryset(self):
        qs = super(ScholarshipDirectoryView, self).get_queryset()
        return qs.filter(experience_level__exact=self.kwargs['exp'])