SQL ALCHEMY -根据父级中指定的条件查询子级

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

Base = declarative_base()

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

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

# 创建数据库引擎和会话
engine = create_engine('sqlite:///example.db')
Session = sessionmaker(bind=engine)
session = Session()

# 查询父级名称为 'Parent1' 的所有子级
parent_name = 'Parent1'
parent = session.query(Parent).filter(Parent.name == parent_name).first()
if parent:
    children = parent.children
    for child in children:
        print(child.name)
else:
    print(f"No parent found with name: {parent_name}")