Question

使用属性和混合变换器在遍历以下树时有所帮助;

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

我有一个简单的对象

catlist = db.session.query(Category).filter_by(id=1).all()

如何遍历所有树以获得具有可变树深度的子项即祖先 - ＆gt;父母 - ＆gt;儿童 - ＆gt;子儿？
我如何只获得一个子子对象？

对于Feed模型，我如何遍历祖先树，以及如何只获取顶级祖先节点？

这是我到目前为止所做的，除了产生错误的属性对象

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

Answer 1

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

@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)

调用：

all_children = Category.get_children_list(1).all()

查询将如下：

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;```

Sqlachemy，通过关系递送儿童和祖先

1 个答案: