SQLAlchemy：树节点中的递归混合属性

Question

我有一个自引用的（树形）节点，想要根据计算出来的属性 uuid_path 和 name_path 进行过滤和排序：

class Node (db.Model):
    id = db.Column (db.Integer, db.Sequence ('node_id_seq'), primary_key=True)

    ###########################################################################

    root_id = db.Column (db.Integer, db.ForeignKey (id, ondelete='CASCADE'),
        index=True)

    nodes = db.relationship ('Node',
        cascade='all, delete-orphan', lazy='dynamic',
        primaryjoin='Node.id==Node.root_id',
        backref=db.backref ('root', remote_side=id))

    ###########################################################################

    _uuid = db.Column (db.String (36), nullable=False, index=True, unique=True,
        name = 'uuid')
    _name = db.Column (db.Unicode (256), nullable=False, index=True,
        name = 'name')

    ###########################################################################

    @hybrid_property
    def uuid (self):
        return self._uuid

    @hybrid_property
    def name (self):
        return self._name
    @name.setter
    def name (self, value):
        self._name = value

    ###########################################################################

    def __init__ (self, name, root, mime=None, uuid=None):

        self.root = root
        self._uuid = uuid if uuid else str (uuid_random ())
        self._name = unicode (name) if name is not None else None

    def __repr__ (self):

        return u'<Node@%x: %s>' % (self.id if self.id else 0, self._name)

    ###########################################################################

    @hybrid_property
    def uuid_path (self):
        node, path = self, []
        while node:

            path.insert (0, node.uuid)
            node = node.root

        return os.path.sep.join (path)

    @hybrid_property
    def name_path (self):
        node, path = self, []
        while node:

            path.insert (0, node.name)
            node = node.root

        return os.path.sep.join (path)

    ###########################################################################

如果我得到一个 Node 实例 subnode，然后执行比如 subnode.name_path，我会得到正确的结果，比如 root/subnode。但是如果我尝试使用 Node.name_path（来进行过滤或排序），SQLAlchemy 就会报错：

Neither 'InstrumentedAttribute' object nor 'Comparator' object associated with Node.root has an attribute 'name'.

我很确定我需要引入一些东西，比如：

class Node (db.Model):

    @hybrid_property
    def name_path (self):
        node, path = self, []
        while node:

            path.insert (0, node.name)
            node = node.root

        return os.path.sep.join (path)

    @name_path.expression
    def name_path (cls):
        ## Recursive SQL expression??

但我在定义 @name_path.expression（或者 @uuid_path.expression）时遇到了困难；它应该以某种方式告诉 SQL 从根节点到当前节点的路径。

我不明白的是，为什么需要这样做，因为我已经告诉 SQLAlchemy 逐步计算路径值了。谢谢你的帮助。

Answer 1

为了完整性，我把zzzeek在他这个链接中的反馈也加上了：https://gist.github.com/4625858

from sqlalchemy.sql.expression import ColumnElement ## !!
from sqlalchemy.ext.hybrid import hybrid_property
from sqlalchemy.ext.compiler import compiles
from sqlalchemy import inspect

class UuidPathColumn (ColumnElement):
    def __init__(self, entity):
        insp = inspect (entity)
        self.entity = insp.selectable

@compiles (UuidPathColumn)
def compile_uuid_path_column (element, compiler, **kwargs):
    return "%s.uuid_path" % compiler.process (element.entity, ashint=True)

class NamePathColumn (ColumnElement):
    def __init__(self, entity):
        insp = inspect (entity)
        self.entity = insp.selectable

@compiles (NamePathColumn)
def compile_name_path_column (element, compiler, **kwargs):
    return "%s.name_path" % compiler.process (element.entity, ashint=True)

要让这个功能正常工作，使用ColumnElement（而不是ColumnClause）是很重要的。相关的代码可以在以下链接找到：node.py、name_path和uuid_path。这些内容是用SQLAlchemy 0.8和PostgreSQL 9.2实现的。

Answer 2

经过一番调整PostgreSQL和SQLAlchemy，我觉得我找到了一个解决方案：(1) 首先，我需要把查询写成一个SQL函数，(2) 其次，提供正确的SQLAlchemy连接方式：

SQL部分使用了一个叫WITH RECURSIVE的CTE：

CREATE OR REPLACE FUNCTION name_path(node)
  RETURNS text AS
$BODY$

WITH RECURSIVE graph (id, root_id, id_path, name_path) AS (
    SELECT n.id, n.root_id, ARRAY[n.id], ARRAY[n.name]
    FROM node n
UNION
    SELECT n.id, n.root_id, id_path||ARRAY[n.id], name_path||ARRAY[n.name]
    FROM node n, graph g
    WHERE n.root_id = g.id)

SELECT array_to_string (g.name_path, '/','.*')
FROM graph g
WHERE (g.id_path[1] = $1.base_id OR g.root_id IS NULL)
AND (g.id = $1.id)

$BODY$
  LANGUAGE sql STABLE
  COST 100;
ALTER FUNCTION name_path(node)
  OWNER TO webed;

而SQLAlchemy的部分看起来是这样的：

class NamePathColumn (ColumnClause):
    pass

@compiles (NamePathColumn)
def compile_name_path_column (element, compiler, **kwargs):
    return 'node.name_path' ## something missing?

还有

class Node (db.Model):

    def get_path (self, field):

        @cache.version (key=[self.uuid, 'path', field])
        def cached_path (self, field):

            if self.root:
                return self.root.get_path (field) + [getattr (self, field)]
            else:
                return [getattr (self, field)]

        if field == 'uuid':
            return cached_path (self, field)
        else:
            return cached_path.uncached (self, field)

    @hybrid_property
    def name_path (self):
        return os.path.sep.join (self.get_path (field='name'))

    @name_path.expression
    def name_path (cls):
        return NamePathColumn (cls)

如果我在纯Python的环境下，我会避免直接访问数据库中的Node.name_path，不过如果我这样做可能会更快。我现在唯一不太确定的是，在compile_name_path_column中，我没有考虑到任何element, compiler, **kwargs这些参数，这让我有点怀疑。

我是在折腾了大约1.5天的SA和PG之后才想出来这个，所以很可能还有改进的空间。如果对这个方法有任何意见，我会非常感激。谢谢。