在SQLalchemy中的左联接上过滤 [英] Filtering on a left join in SQLalchemy

问题描述

我要使用SQLalchemy执行左外部联接，并过滤掉联接表中确实有匹配项的行.

Using SQLalchemy I want to perform a left outer join and filter out rows that DO have a match in the joined table.

我正在发送推送通知，所以我有一个Notification表.这意味着我还有一个ExpiredDeviceId表来存储不再有效的device_id. (我不想删除受影响的通知，因为用户以后可能会重新安装该应用，此时通知应根据Apple的文档恢复.)

I'm sending push notifications, so I have a Notification table. This means I also have a ExpiredDeviceId table to store device_ids that are no longer valid. (I don't want to just delete the affected notifications as the user might later re-install the app, at which point the notifications should resume according to Apple's docs.)

CREATE TABLE Notification (device_id TEXT, time DATETIME);
CREATE TABLE ExpiredDeviceId (device_id TEXT PRIMARY KEY, expiration_time DATETIME);

注意:每个device_id可能有多个通知.每个设备没有设备"表.

因此，当执行SELECT FROM Notification时，我应该进行相应的过滤.我可以用SQL做到:

So when doing SELECT FROM Notification I should filter accordingly. I can do it in SQL:

SELECT * FROM Notification 
    LEFT OUTER JOIN ExpiredDeviceId 
    ON Notification.device_id = ExpiredDeviceId.device_id
WHERE expiration_time IS NULL

但是我如何在SQLalchemy中做到这一点?

But how can I do it in SQLalchemy?

sess.query(
    Notification, 
    ExpiredDeviceId
).outerjoin(
    (ExpiredDeviceId, Notification.device_id == ExpiredDeviceId.device_id)
).filter(
    ???
)

或者，我可以使用device_id NOT IN (SELECT device_id FROM ExpiredDeviceId)子句来执行此操作，但这似乎效率较低.

Alternately I could do this with a device_id NOT IN (SELECT device_id FROM ExpiredDeviceId) clause, but that seems way less efficient.

推荐答案

是否需要在元组中拉回ExpiredDeviceId?如果您不这样做(即，您只是关心实时的device_ids)，那么您就不能这样做:

Do you need to be pulling back the ExpiredDeviceId in a tuple? If you don't (i.e. you just care about live device_ids), then can't you just do:

sess.query(
    Notification
).outerjoin(
    (ExpiredDeviceId, Notification.device_id == ExpiredDeviceId.device_id)
).filter(
    ExpiredDeviceId.expiration_time == None
)

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