Django的：集團通過澄清

我有一個views.py看起來是這樣的：Django的：集團通過澄清

category = Category.objects.get(id=categoryid)  
#posts = get_list_or_404(Post, category=categoryid) 
posts = Post.objects.filter(category=categoryid,is_comment=0).order_by('-published') 
all_posts = [] 

for p in posts: 
    all_posts += [(x.id, x.marked_read_on, p, \ 
        Unread.objects.filter(marked_read_on__isnull=True, \ 
              user=request.user,comment__post=p.id).count(), \ 
        Unread.objects.filter(user=request.user,comment__post=p.id).count(), \ 
        #1 
       ) \ 
       #for x in p.unread_set.filter(post=p.id)] 
       for x in p.unread_set.filter(post=p.id).annotate(Count('post')).order_by('post')]

對於for x in p.unread_set.filter(post=p.id).annotate(Count('post')).order_by('post')]我得到下面的SQL查詢：（沒有錯誤）

SELECT `message_unread`.`id`, `message_unread`.`user_id`, `message_unread`.`post_id`, `message_unread`.`comment_id`, `message_unread`.`category_id`, `message_unread`.`marked_unread_on`, `message_unread`.`marked_read_on`, COUNT(`message_unread`.`post_id`) AS `post__count` FROM `message_unread` WHERE (`message_unread`.`post_id` = 4 AND `message_unread`.`post_id` = 4) GROUP BY `message_unread`.`id` ORDER BY `message_unread`.`post_id` ASC

我只是想這部分： GROUP BY message_unread。 id

成爲： GROUP BY message_unread。 post_id

我只是想通過post_id對結果進行分組。僅供參考，應用程序的名稱是「消息」，這就是爲什麼查詢具有「message_」前綴。

我該怎麼做？

更新，我的models.py如下：

class Category(models.Model): 
    name = models.CharField(max_length=120) 
    group = models.ForeignKey(Group) 

    def __unicode__(self): 
     return self.name 


class Post(models.Model): 
    title = models.CharField(max_length=250) 
    #slug = models.SlugField(max_length=250, unique=True) 
    body = models.TextField() 
    published = models.DateTimeField(default=datetime.now) 
    category = models.ForeignKey(Category) 
    group = models.ForeignKey(Group) 
    user = models.ForeignKey(User) 
    is_comment = models.BooleanField(default=0) 

    def __unicode__(self): 
     return self.title 

class Comment(models.Model): 
    post = models.ForeignKey(Post, related_name='comment_parent') 
    comment = models.ForeignKey(Post) 

    def __unicode__(self): 
     return comment.title 

class Unread(models.Model): 
    user = models.ForeignKey(User) 
    post = models.ForeignKey(Post) 
    comment = models.ForeignKey(Comment, null=True, blank=True) 
    category = models.ForeignKey(Category) 
    marked_unread_on = models.DateTimeField(null=True, blank=True) 
    marked_read_on = models.DateTimeField(null=True, blank=True)

來源

2011-03-03 wenbert

下面是unread.objects.filter(post=p.id).values(message_unread.id).order_by().annotate(Count('post'))

>>> from django.db.models import Count 
>>> m=Unread.objects.filter(post__id='1').values('id').order_by().annotate(Count('post')) 
>>> m 
[{'id': 1L, 'post__count': 1}] 
>>> m.values() 
[{'marked_read_on': datetime.datetime(2011, 3, 3, 22, 3, 33), 'user_id': 1L, 'comment_id': 1L, 'post_id': 1L, 'marked_unread_on': datetime.datetime(2011, 3, 3, 22, 3, 29), 'category_id': 1L, 'id': 1L, 'post__count': 1}]

以下查詢

>>> print m.query 
SELECT `mytest_unread`.`id`, COUNT(`mytest_unread`.`post_id`) AS `post__count` FROM `mytest_unread` WHERE `mytest_unread`.`post_id` = 1 GROUP BY `mytest_unread`.`id`, `mytest_unread`.`id` ORDER BY NULL

來源

2011-03-03 16:06:31 sush

招呼舒尚特，感謝，但它組查詢仍然不適合我。我已經更新了我的文章，幷包含了我的模型，以澄清字段關係。 – wenbert 2011-03-03 16:14:58

>>> print m.query SELECT'mytest_unread'.'''，COUNT（'mytest_unread'.'post_id'）AS'post__count' FROM'mytest_unread' WHERE'mytest_unread'.'post_id' = 1 GROUP BY'mytest_unread '.'id'，'mytest_unread'.'id' ORDER BY NULL – sush 2011-03-03 16:52:00

Django的：集團通過澄清

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