保存模型实例时出现KeyError。Django

Question

在尝试保存模型实例时出现了KeyError错误。这意味着在保存实例时，它需要对post_save信号做出反应...

代码：

from django.db.models.signals import post_save

class PlaylistEntry(models.Model):
    playlist=models.ForeignKey(Playlist)
    media=models.ForeignKey(Media)
    order=models.PositiveIntegerField(default=9000000, editable=False)
    added=models.DateTimeField(default=datetime.datetime.now(),editable=False )

def playlist_entry_changed(sender, instance, **kwargs):
        entrys=PlaylistEntry.objects.filter(playlist=instance.playlist).order_by('order')
        entrys[0].save()

post_save.connect(playlist_entry_changed, PlaylistEntry)

错误：

Exception Type: KeyError at /admin/playlist/playlistentry/add/
Exception Value: 38539456

Answer 1

如果你想实现排序功能，可以看看这个代码片段！

Answer 2

根据你的评论，你想要做的是更新排序。与其使用信号，不如重写保存方法。

def save(self, *args, **kwargs):
    # Only do this if it's the first time we're saving.
    if not self.id:
        entries = PlaylistEntry.objects.order_by('-order')
        try:
            self.order = entries[0].order + 1
        except IndexError:
            # we don't have any PlaylistEntries yet, so we just start @ 0
            self.order = 0

    super(PlaylistEntry, self).save(*args, **kwargs) 

另外，为什么你的默认排序是9000000呢……难道不应该从0开始吗？