python - SQLAlchemy 左联接 WHERE 子句被转换为零和一

Question

你好，

我有以下 SQL，我正在将其转换为 SQLAlchemy:

select t1.`order_id`, t1.`status_type`
from `tracking_update` AS t1 LEFT JOIN `tracking_update` AS t2
ON (t1.`order_id` = t2.`order_id` AND t1.`last_updated` < t2.`last_updated`)
where t1.`order_id` = '21757'and t2.`last_updated` IS NULL

SQL 只是返回订单 ID 21757 的最新跟踪更新。我通过向同一个表执行左连接来完成此操作。为了做到这一点，我首先为表格设置了别名:

        tUAlias1 = aliased(TrackingUpdate)
        tUalias2 = aliased(TrackingUpdate)

到目前为止，这是我转换到 SQLAlchemy 的结果:

tracking_updates = db.session.query(tUAlias1.order_id, tUAlias1.status_type).\
            outerjoin(tUalias2, (tUAlias1.order_id == tUalias2.order_id) & (tUAlias1.last_updated < tUalias2.last_updated)).\
            filter(and_(tUAlias1.order_id == '21757', tUalias2.last_updated is None))

这是通过日志在服务器上执行的 SQLAlchemy 代码的结果:

SELECT tracking_update_1.order_id AS tracking_update_1_order_id, tracking_update_1.status_type AS tracking_update_1_status_type 
FROM tracking_update AS tracking_update_1 LEFT OUTER JOIN tracking_update AS tracking_update_2 ON tracking_update_1.order_id = tracking_update_2.order_id AND tracking_update_1.last_updated < tracking_update_2.last_updated 
WHERE 0 = 1

如您所见，过滤器(WHERE 子句)现在为 0 = 1。

现在，如果我删除 and_ 语句并像这样尝试两个过滤器:

tracking_updates = db.session.query(tUAlias1.order_id, tUAlias1.status_type).\
            outerjoin(tUalias2, (tUAlias1.order_id == tUalias2.order_id) & (tUAlias1.last_updated < tUalias2.last_updated)).\
            filter(tUAlias1.order_id == '21757').filter(tUalias2.last_updated is None)

我收到了相同的结果。我知道 SQL 本身很好，因为我可以通过 MySQL Workbench 毫无问题地运行它。

当SQL直接运行时，会收到如下信息

order ID | Status
21757      D

此外，如果我删除 tUalias2.last_updated is None，我实际上会收到一些结果，但它们不正确。这是 SQL 日志:

Python代码

tracking_updates = db.session.query(tUAlias1.order_id, tUAlias1.status_type).\
            outerjoin(tUalias2, (tUAlias1.order_id == tUalias2.order_id) & (tUAlias1.last_updated < tUalias2.last_updated)).\
            filter(tUAlias1.order_id == '21757')

SQLAlchemy 运行:

SELECT tracking_update_1.order_id AS tracking_update_1_order_id, tracking_update_1.status_type AS tracking_update_1_status_type 
FROM tracking_update AS tracking_update_1 LEFT OUTER JOIN tracking_update AS tracking_update_2 ON tracking_update_1.order_id = tracking_update_2.order_id AND tracking_update_1.last_updated < tracking_update_2.last_updated 
WHERE tracking_update_1.order_id = '21757'

有什么想法吗？

Answer 1

你好，

我想通了

Python 的 'is' 运算符不能很好地与 SQLAlchemy 搭配使用

由于以下 S/O 问题，我发现了这一点:

Selecting Null values SQLAlchemy

我已将我的查询更新为以下内容:

tracking_updates = db.session.query(tUAlias1.order_id, tUAlias1.status_type).\
            outerjoin(tUalias2, (tUAlias1.order_id == tUalias2.order_id) & (tUAlias1.last_updated < tUalias2.last_updated)).\
            filter(tUAlias1.order_id == '21757').filter(tUalias2.last_updated == None)