Sqlachemy,递归获取有关系的孩子和祖先

afi*_*num 0 python sqlalchemy flask flask-sqlalchemy

使用属性和混合变压器对遍历以下树有一点帮助;

class Category(db.Model):
    __tablename__ = 'category'
    id = db.Column(db.Integer, primary_key=True)
    parent = db.Column(db.Integer, db.ForeignKey('category.id'), nullable=True)
    name = db.Column(db.String(400), index=True, unique=True)

    children = db.relationship("Category", cascade='all, delete-orphan', backref=db.backref("child", remote_side=[id]))
    parents = db.relationship("Category", cascade='all', backref=db.backref("back", remote_side=[id]))
    entries = db.relationship("Feeds", backref='entry', lazy='dynamic')


class Feeds(db.Model):
    __tablename__ = 'feeds'
    id = db.Column(db.Integer, primary_key=True)
    category_id = db.Column(db.Integer, db.ForeignKey('category.id'))
    name = (db.String(400), index=True, unique=True)

    @property
     def parents(self):
         allparents=[]
         p = self.children
         while p:
             allparents.append(p)
             p = p.children
         return allparents
Run Code Online (Sandbox Code Playgroud)

我有一个简单的对象

catlist = db.session.query(Category).filter_by(id=1).all()
Run Code Online (Sandbox Code Playgroud)
  1. 如何遍历所有树以获得具有可变树深度的子级,即祖先 -> 父级 -> 子级 -> 子子级?

  2. 我如何只获得一个子子对象?

Feed模型也是一样,如何遍历祖先树,如何只获取最顶层的祖先节点?

这是我到目前为止所做的,除了生成错误的属性对象外,它似乎效果不佳

catlist = db.session.query(Category).filter_by(id=1).all()
for cat in catlist:
    cat[0].children



File "/home/afidegnum/PycharmProjects/store/core/model.py", line 45, in children
    p = self.children
RuntimeError: maximum recursion depth exceeded
Run Code Online (Sandbox Code Playgroud)

ant*_*uan 7

使用 Postgresql,您可以使用递归查询。在您的情况下,您可以使用以下方法:

@staticmethod
def get_parents_list(category_id):
    beginning_getter = Session.query(Category).\
        filter(Category.id == category_id).cte(name='parent_for', recursive=True)
    with_recursive = beginning_getter.union_all(
        Session.query(Category).filter(Category.id == beginning_getter.c.parent_id)
        )
    return Session.query(with_recursive)

@staticmethod
def get_children_list(category_id):
    beginning_getter = Sesion.query(Category).\
            filter(Category.id == category_id).cte(name='children_for', recursive=True)
    with_recursive = beginning_getter.union_all(
            Session.query(Category).filter(Category.parent_id == beginning_getter.c.id)
        )
    return Session.query(with_recursive)
Run Code Online (Sandbox Code Playgroud)

调用:

all_children = Category.get_children_list(1).all()
Run Code Online (Sandbox Code Playgroud)

查询将是这样的:

WITH RECURSIVE children_for(id, name, parent) AS
(SELECT id, name, parent
FROM categories
WHERE category.id = 1 UNION ALL id, name, parent
FROM categories, children_for
WHERE categories.parent = children_for.id)
 SELECT children_for.id, children_for.name, children_for.parent
FROM children_for;```
Run Code Online (Sandbox Code Playgroud)