Django - 评级模型示例DetailView模板 [英] Django - Rating Model Example DetailView Template

问题描述

这是上下文：
我有用户，视频，主题，标准和评分

Here is the context: I have users, videos, topics, criterias and ratings

• 视频有一个主题


• 主题具有标准


• 用户可以为给定的主题创建视频


• 用户可以为有关主题给出的标准评价一个视频。

• A video has a topic
• A topic has criterias
• A user can create a video for a given topic
• A user can rate a video on the criterias given for the concerned topic.

这是我的模型为这个目的：

Here is my model for that purpose:

RATE_CHOICES = zip( range(1,5), range(1,5) )

class VideoCrit(models.Model):
  """Criteria to rate videos on.
  Can be multiple for each Topic of Video"""
  name = models.CharField(max_length=50)
  def __unicode__(self):
    return self.name
  class Meta:
    verbose_name = 'Video Criteria'

class VideoTopic(models.Model):
  name = models.CharField(max_length=50)
  descr = models.TextField(blank=True, null=True)
  crits = models.ManyToManyField(VideoCrit,
    help_text='Criterias to rate the videos',
    blank=True, null=True,
  )
  def __unicode__(self):
    return self.name
  class Meta:
    verbose_name = 'Video Topic'

class VideoFile(models.Model):
  """Uploadable by users to be rated and commented"""
  name = models.CharField(max_length=50)
  descr = models.TextField(blank=True, null=True)
  file = models.FileField(upload_to='videos')
  topic = models.ForeignKey(VideoTopic)
  def __unicode__(self):
    return self.name
  class Meta:
    verbose_name = 'Chatter Tube'

class VideoRate(models.Model):
  """Users can Rate each Video on the criterias defined for the topic"""
  user = models.ForeignKey(User)
  video = models.ForeignKey(VideoFile)
  crit = models.ForeignKey(VideoCrit)
  rate = models.DecimalField(max_digits=2, decimal_places=1, choices=RATE_CHOICES)
  class Meta:
    unique_together = (('user', 'video', 'crit'),)
    verbose_name = 'Tube Rating'

可以吗？

如果是，从给定VideoFile的模板（基于VideoFile类的DetailView）中，以下是模板的有趣部分

If yes, from a template (DetailView based on the VideoFile class) for a given VideoFile, here is the interesting part of the template

  <div id="rating">
    <ul>
  {% for crit in videofile.topic.crits.all %}
      <li>
        <div class="rateit"
          crit-id="{{ crit.id }}"></div>
        {{ crit.name }}
      </li>
  {% endfor %}
    </ul>
  </div>

URLconf&查看

URLconf & View

#urlconf
#...
  (r'viewtube/(?P<pk>\d+)$', VideoFileDetailView.as_view()),
#...

#view
class VideoFileDetailView(DetailView):
  model = VideoFile
  def get_context_data(self, **kwargs):
    context = super(VideoFileDetailView, self).get_context_data(**kwargs)
#    context['rates'] = VideoRate.objects.filter(video=11)
    return context

如何访问当前记录的评分用户为当前视频？

How can I access to the ratings of the current logged user for the current video ?

推荐答案

更新：获取当前视频的当前登录用户的所有评分

UPDATE: To get all the ratings for the currently logged in user for the current video

# in Views.py 

video = VideoFile.objects.get(pk=video_id) #video_id is parameter sent from url
user_ratings = VideoRate.objects.filter(user=request.user).filter(video=video)

# in template
<ul>
{% for rating in user_ratings %}
    <li>{{ rating.crit.name }}: {{ rating.rate }}</li>
{% endfor %}
</ul>

以前：

你应该能够访问登录用户的评分，使用这样的效果：

You should be able to access the ratings of the logged in user using something to this effect:

user.videorate_set.all

您可以在模板中显示给定用户的所有评分，如下所示：

You can then display all of the ratings for a given user in your template as follows:

{% for rating in user.videorate_set.all %}
    {{ rating.video }} {{ ratings.rate }}
{% endfor %}

这篇关于Django - 评级模型示例DetailView模板的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！