Question

有没有办法简化这个工作代码？这段代码为一个对象获取所有不同的投票类型，有20种可能，并计算每种类型。我不想写原始的SQL但使用orm。这有点棘手，因为我在模型中使用通用外键。

def get_object_votes(self, obj):
    """ 
    Get a dictionary mapping vote to votecount
    """
    ctype = ContentType.objects.get_for_model(obj)

    cursor = connection.cursor()
    cursor.execute("""
       SELECT v.vote , COUNT(*)
       FROM votes v
       WHERE %d = v.object_id AND %d = v.content_type_id
       GROUP BY 1
       ORDER BY 1 """ % ( obj.id, ctype.id )
    )
    votes = {}

    for row in cursor.fetchall():
       votes[row[0]] = row[1]

    return votes

使用

的模型

class Vote(models.Model):
    user = models.ForeignKey(User)

    content_type = models.ForeignKey(ContentType)
    object_id = models.PositiveIntegerField()
    payload = generic.GenericForeignKey('content_type', 'object_id')

    vote = models.IntegerField(choices = possible_votes.items() )


class Issue(models.Model):
    title = models.CharField( blank=True, max_length=200)

Answer 1

以下代码为我做了诀窍！

def get_object_votes(self, obj, all=False):
    """
    Get a dictionary mapping vote to votecount
    """
    object_id = obj._get_pk_val()
    ctype = ContentType.objects.get_for_model(obj)
    queryset = self.filter(content_type=ctype, object_id=object_id)

    if not all:
        queryset = queryset.filter(is_archived=False) # only pick active votes

    queryset = queryset.values('vote')
    queryset = queryset.annotate(vcount=Count("vote")).order_by()

    votes = {}

    for count in queryset:
        votes[count['vote']] = count['vcount']

    return votes

Answer 2

是的，绝对使用ORM。你应该在你的模型中做到这一点：

class Obj(models.Model):
    #whatever the object has

class Vote(models.Model):
    obj = models.ForeignKey(Obj) #this ties a vote to its object

然后要从对象获得所有投票，请将这些Django调用放在您的一个视图函数中：

obj = Obj.objects.get(id=#the id)
votes = obj.vote_set.all()

从那里可以很容易地看出如何计算它们（得到名单的长度称为投票）。

我建议从文档中阅读多对一关系，这非常方便。

http://www.djangoproject.com/documentation/models/many_to_one/

将原始sql映射到django orm

2 个答案: