Question

我有一个User模型，每个用户都有另一个用户作为父级。现在，为了获得子节点数（属于给定模型实例的用户），我已经定义了这个属性：

class User( object ):

    @property
    def childsCount( self ):
        return object_session(self).scalar(
            select([func.count(User.users_id)]).where(User.parent_id==self.users_id)
        )

......哪作得好。我不知道的是如何获得孙子的数量？甚至是爷爷们。

有什么想法吗？

Answer 1

使用aliased编写更深层WHERE子句。事实上，你可以做得更通用：

@property
def childrenCount(self):
    return self.count_children(0)

@property
def grandchildrenCount(self):
    return self.count_children(1)

@property
def grandgrandchildrenCount(self):
    return self.count_children(2)

def count_children(self, level=0):
    a = [aliased(User) for _ in range(level + 1)]
    qry = select([func.count(a[0].users_id)]).where(a[-1].parent_id==self.users_id)
    # insert all the intermediate JOINs
    for _i in range(level):
        qry = qry.where(a[_i].parent_id == a[_i+1].users_id)
    return Session.object_session(self).scalar(qry)

虽然它看起来有点神秘，但它真正做的是如下所示（为每个更深层次添加一个alias和where子句）：

@property
def children1Count(self):
    a0 = aliased(User)
    qry = select([func.count(a0.users_id)]).where(a0.parent_id==self.users_id)
    return Session.object_session(self).scalar(qry)

@property
def children2Count(self):
    a0 = aliased(User)
    a1 = aliased(User)
    qry = select([func.count(a0.users_id)]).where(a0.parent_id==a1.users_id).where(a1.parent_id==self.users_id)
    return Session.object_session(self).scalar(qry)

@property
def children3Count(self):
    a0 = aliased(User)
    a1 = aliased(User)
    a2 = aliased(User)
    qry = select([func.count(a0.users_id)]).where(a0.parent_id==a1.users_id).where(a1.parent_id==a2.users_id).where(a2.parent_id==self.users_id)
    return Session.object_session(self).scalar(qry)

仅限第一级，您可以使用with_parent实际获得更好的查询：

@property
def childrenCount(self):
    return Session.object_session(self).query(User).with_parent(self).count()

SQLAlchemy自引用表获取孙子的数量

1 个答案: