python - SQLAlchemy和Python-返回ID的upsert语句

我正在写一个sqlalchemy upsert语句，我想知道是否有办法可以从数据库取回ID？
到目前为止，这是我的代码：

def handle_events(self, events):
    agg = []
    agg.extend(self._get_dicts(events))
    if not agg:
        return
    with DB.session(writable=True) as session:
        response = session.execute(self._generate_upsert_stmnt(agg))


def _generate_upsert_stmnt(self, items):
    model_class = models.MyModel
    table = model_class.__table__
    insert_statement = sa.dialects.postgresql.insert(table, items)
    upsert_statement = insert_statement.on_conflict_do_update(
        index_elements=[table.c['id']],
        set_={c.name: c for c in insert_statement.excluded if c.name != ["id", "my_model", "my_model_id"]}
    )
    return upsert_statement

此代码段中的会话是来自sqlalchemy的api（包装在上下文管理器中）的sessionmaker调用返回的对象。有没有办法使此upsert语句返回所有id的背面（插入和更新）？如果是这样，该代码将如何显示？
提前致谢。 ps。如果很重要，数据库就是postgres。

最佳答案

最终，我在upsert语句中使用了.returning(<column>)，如下所示：

def _generate_upsert_stmnt(self, items):
    model_class = models.MyModel
    table = model_class.__table__
    insert_statement = sa.dialects.postgresql.insert(table, items)
    upsert_statement = insert_statement.on_conflict_do_update(
        index_elements=[table.c['id']],
        set_={c.name: c for c in insert_statement.excluded if c.name !=
        ["id", "my_model", "my_model_id"]}).returning(table.c['id'])
    return upsert_statement

并像这样提取数据：

    def _extract_ids_from_resultproxy(self, result):
        table = models.MyModel.__table__
        return [row[table.c.id] for row in result]