Sqlalchemy: Query -当所有子节点都满足不等条件时，仅获取父节点

from sqlalchemy import Column, Integer, String, ForeignKey
from sqlalchemy.orm import relationship
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    children = relationship('Child')

class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    parent_id = Column(Integer, ForeignKey('parent.id'))

from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker

engine = create_engine('数据库连接字符串')
Session = sessionmaker(bind=engine)
session = Session()

# 构建子查询，获取所有满足不等条件的子节点的父节点ID
subquery = session.query(Child.parent_id).filter(Child.name != '条件')

# 查询所有子节点都满足不等条件的父节点
parents = session.query(Parent).filter(Parent.id.in_(subquery)).all()