Question

这里我定义了一个模型来在'models.py'中创建一个feed实例：

class StreamItem(models.Model):
  content_type = models.ForeignKey(ContentType)
  object_id = models.PositiveIntegerField()
  pub_date = models.DateTimeField()

  content_object = generic.GenericForeignKey('content_type', 'object_id')

  def get_rendered_html(self):
    template_name = 'streams/stream_item_%s.html' % (self.content_type.name)
    return render_to_string(template_name, { 'object': self.content_object })

def create_stream_item(sender, instance, signal, *args, **kwargs):
    # Check to see if the object was just created for the first time
    if 'created' in kwargs:
        if kwargs['created']:
            create = True

            # Get the instance's content type
            ctype = ContentType.objects.get_for_model(instance)

            pub_date = instance.pub_date

            if create:
                si = StreamItem.objects.get_or_create(content_type=ctype, object_id=instance.id, pub_date=pub_date)

# Send a signal on post_save for each of these models
for modelname in [Fest, College, Event]: 
    my_signal = dispatch.Signal()      
    my_signal.connect(create_stream_item, sender=modelname)

如果我从管理站点创建 StreamItem 对象，那么我已经创建了用于获取正常工作的视图的视图。但是，信号部分不仅仅是工作。我只是在学习它，所以，我不明白我错在哪里。请帮忙。

Answer 1

使用post_save信号：

from django.db.models.signals import post_save

for model in [Fest, College, Event]:
    post_save.connect(create_stream_item, sender=model)

使用Django中的信号创建feed

1 个答案: