sqlalchemy限制从第一个连接返回的结果，并使后续连接仅对有限的结果进行连接

from sqlalchemy import create_engine, Table, MetaData, select, func

engine = create_engine('sqlite:///:memory:')  # 示例使用 SQLite 内存数据库
metadata = MetaData()

# 定义表结构
TableA = Table('table_a', metadata,
               Column('id', Integer, primary_key=True),
               Column('value', String))
TableB = Table('table_b', metadata,
               Column('id', Integer, primary_key=True),
               Column('a_id', Integer, ForeignKey('table_a.id')),
               Column('data', String))

# 创建表
metadata.create_all(engine)

# 插入示例数据
with engine.connect() as conn:
    conn.execute(TableA.insert(), [{'value': f'value{i}'} for i in range(100)])
    conn.execute(TableB.insert(), [{'a_id': i % 100, 'data': f'data{i}'} for i in range(100)])

# 使用子查询限制结果
subquery = select(TableA.c.id).limit(10).subquery()
stmt = select(TableA, TableB).join(TableB, TableA.c.id == TableB.c.a_id).where(TableA.c.id.in_(subquery))

with engine.connect() as conn:
    result = conn.execute(stmt)
    for row in result:
        print(row)