在Django 1.7中动态过滤ListView CBV [英] Dynamically filter ListView CBV in Django 1.7

问题描述

我已阅读关于动态过滤的ListView的官方文档，但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，我们称之为奖学金：

pre> 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'）


我有一个页面显示所有这些奖学金，使用ListView：

视图。 py

  from django.views.generic import ListView 
 from .models import奖学金
 
 
 class ScholarshipDirectoryView（ListView）：
 model =奖学金
 template_name ='scholarship-directory.html'
  

urls.py

 从django.conf.urls导入模式， url 
 
 from .views import ScholarshipDirectoryView 
 
 urlpatterns = patterns（''，
 url（r'^ $'，ScholarshipDirectoryView.as_view（），name ='奖学金目录'），
）
  

我试图在主页上生成链接该网站将返回此ListView的过滤版本。例如，如果有人点击为研究生颁发奖学金链接，那么只会显示 experience_level ='G'的奖学金。

我没有问题通过shell返回此查询器 - > 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。

解决方案

首先，您需要更改urls.py，以便将经验作为参数传递。这样的事情：

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

（以上将返回404 if / A或/ S或/ G未通过）

现在，在CBV的 kwargs 属性中我们将有一个名为 exp 的kwarg，可以由 get_queryset 方法用于按经验级别进行过滤。

 
 class ScholarshipDirectoryView（ListView）：
 model =奖学金
 template_name ='scholarship-directory.html'
 
 def get_queryset（self）：
 qs = super（ScholarshipDirectoryView，self）.get_queryset（）
 return qs.filter（experience_level__exact = self.kwargs ['exp']）
 

I've read the official documentation on dynamically filtering ListView, but am still confused about how to actually use it.

I currently have a simple model, let's call it 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')

I have a page where I'm showing all of these scholarships, using ListView:

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'),
)

I'm trying to generate links on the home page of the site that will return filtered versions of this ListView. For example, if someone clicks on a "show scholarships for graduate students" link, only scholarships with experience_level='G' will be shown.

I have no problem returning this queryset via the shell -> Scholarship.objects.filter(experience_level__exact='G')

I'm just unsure about how to dynamically filter the ListView via a dropdown or URL. Not looking to use a plugin, but rather understand how dynamically querying/filtering works in Django.

解决方案

First of all you need to change your urls.py so that it'll pass the experience as a parameter. Something like this:

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

(the above will return 404 if /A or /S or /G is not passed)

Now, in kwargs attribute of the CBV we will have a kwarg named exp which can be used by the get_queryset method to filter by experience level.

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'])

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