Question

我有以下自引用（树）节点，并希望按计算的属性uuid_path和name_path进行过滤/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)

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

如果我获得Node个实例subnode并执行例如subnode.name_path然后我得到正确的信息，例如root/subnode。但是如果我尝试使用Node.name_path（用于过滤/排序），那么SQLAlchemy会抱怨：

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

我很确定我会介绍类似的内容：

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??

但我很难为@name_path.expression（或@uuid_path.expression）获得正确的定义;它应该以某种方式指示SQL传递从根节点到相关节点的路径。

我不明白为什么这是必需的，因为我告诉SQLAlchemy迭代计算路径值。谢谢你的帮助。

Answer 1

在使用PostgreSQL和SQLAlchemy进行调整之后，我认为我有一个解决方案：（1）首先，我将查询作为SQL中的函数编写，然后（2）第二次提供正确的SQLAlchemy粘合剂：< / p>

SQL部分使用WITH RECURSIVE 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;

并且SQLAlchemy端看起来像这样：

class NamePathColumn (ColumnClause):
    pass

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

和

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方面，我会避免访问数据库上的Node.name_path，但如果我愿意，可能会更快。我唯一不确定的是compile_name_path_column我不考虑任何element, compiler, **kwargs参数，这让我有点怀疑。

我用SA＆amp; amp;修补了大约1.5天后，我就把它煮熟了。 PG，所以很有可能还有改进的余地。我非常感谢w.r.t的任何言论。这种方法。感谢。

Answer 2

为了完整起见，我在他的https://gist.github.com/4625858中加入了zzzeek的反馈意见：

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)

使用ColumnElement（而不是ColumnClause）来实现此功能非常重要;相关代码可在node.py，name_path和uuid_path找到。这个东西已经用SQLAlchemy 0.8和PostgreSQL 9.2实现。

SQLAlchemy：树节点中的递归混合属性

2 个答案: