Python Sqlalchemy筛选器父表和父表的子表

我尝试搜索一些模型，因此我希望显示所有父对象的返回结果，其中父对象名为foo或子对象名为foo

我有一个疑问：

parents = Session.query(Parent).\
            join(Child_s3).\
            filter(Parent.name.ilike("%foo%")).\
            filter(Child_s3.name.ilike("%foo%")).\
            order_by(asc(Product.name))

模型：

class Parent(BaseSO):
    __tablename__ = 'parents'
    id = Column(Integer, primary_key=True)
    name = Column(Unicode(100), nullable=False, unique=True)
    colours = relationship('Child_s3', secondary=Parent_images, backref='Parentc')

class Child_s3(BaseSO):
    __tablename__ = 'children'
    id = Column(Integer, primary_key=True)
    name = Column(Unicode)

Parent_images = Table(
    'Parent_images', BaseSO.metadata,
    Column('parent_id', Integer, ForeignKey('parents.id')),
    Column('child_id', Integer, ForeignKey('children.id'))
)

---

我的查询显示了名为foo的父对象，但没有显示任何父对象，这些父对象也有名为foo的子对象，有人可以帮助构建此查询以搜索两个表中对应的父对象吗？

此代码显示了如何使用显式联接或子查询获得结果：

import sqlalchemy as sa
from sqlalchemy import orm
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()


Parent_images = sa.Table(
    'Parent_images', Base.metadata,
    sa.Column('parent_id', sa.Integer, sa.ForeignKey('parents.id')),
    sa.Column('child_id', sa.Integer, sa.ForeignKey('children.id'))
)


class Parent(Base):
    __tablename__ = 'parents'
    id = sa.Column(sa.Integer, primary_key=True)
    name = sa.Column(sa.Unicode(100), nullable=False, unique=True)
    colours = orm.relationship('Child_s3', secondary=Parent_images, backref='parents')

    def __repr__(self):
        return 'Parent(name=%s)' % self.name

    __str__ = __repr__


class Child_s3(Base):
    __tablename__ = 'children'
    id = sa.Column(sa.Integer, primary_key=True)
    name = sa.Column(sa.Unicode)

    def __repr__(self):
        return 'Child_s3(name=%s)' % self.name

    __str__ = __repr__


if __name__ == '__main__':
    engine = sa.create_engine('sqlite:///')
    Base.metadata.drop_all(engine)
    Base.metadata.create_all(engine)
    Session = orm.sessionmaker(bind=engine)

    session = Session()
    for parent, child in [('boofoo', 'spam'), ('baz', 'foobar'), ('bar', 'quux')]:
        p1 = Parent(name=parent)
        session.add(p1)
        p1.colours.append(Child_s3(name=child))
    session.commit()

    print('Join')
    session = Session()
    q = (session.query(Parent)
                .join(Child_s3, Parent.colours)
                .filter(sa.or_(Parent.name.ilike('%foo%'),
                               Child_s3.name.ilike('%foo%'))))
    for p in q.all():
        print(p, p.colours)
    session.commit()
    print()

    print('Subquery')
    session = Session()
    q = (session.query(Parent)
                .filter(sa.or_(Parent.name.ilike('%foo%'),
                               Parent.colours.any(Child_s3.name.ilike('%foo%')))))
    for p in q.all():
        print(p, p.colours)
    session.commit()
    print()

连接查询

q = (session.query(Parent)
            .join(Child_s3, Parent.colours)
            .filter(sa.or_(Parent.name.ilike('%foo%'),
                           Child_s3.name.ilike('%foo%'))))

生成此SQL

选择parents.id作为parents\u id，parents.name作为parents\u name 从父项将父项图像作为父项图像加入父项图像。id=父项图像加入父项图像。id=父项图像加入子项。id=父项图像加入子项图像。id=父项图像加入子项图像 其中lowerparents.name类似于lower或lowerchildren.name类似于lower 子查询

q = (session.query(Parent)
            .filter(sa.or_(Parent.name.ilike('%foo%'),
                            Parent.colours.any(Child_s3.name.ilike('%foo%')))))

生成以下SQL：

选择parents.id作为parents\u id，parents.name作为parents\u name 来自父母 如果lowerparents.name（如lower或）存在，请选择1 从父图像、子图像 其中parents.id=Parent\u images.Parent\u id和children.id=Parent\u images.child\u id和lowerchildren.name 脚本从示例数据生成以下输出：

参加 Parentname=baz[Child\u s3name=foobar] Parentname=boofoo[Child\u s3name=spam] 子查询 Parentname=boofoo[Child\u s3name=spam] Parentname=baz[Child\u s3name=foobar]