我有几个不同的通知表,我想对所有通知表进行联合,以向用户显示他们的所有通知。然而,工会并没有按应有的方式工作。
Python代码
def _get_notifications_query(self, unconfirmed_only=True):
'''
Return base query to return this users notifications.
@param unconfirmed_only
@return Query object
'''
requests = (
DBSession.query(FriendshipRequestNotification)
.outerjoin(UserFriendshipRequestNotification,
UserFriendshipRequestNotification.notification_id==FriendshipRequestNotification.id)
.filter(UserFriendshipRequestNotification.user_id==self.id))
confirmations = (
DBSession.query(FriendshipConfirmationNotification)
.outerjoin(UserFriendshipConfirmationNotification,
UserFriendshipConfirmationNotification.notification_id==FriendshipConfirmationNotification.id)
.filter(UserFriendshipConfirmationNotification.user_id==self.id))
comments = (
DBSession.query(CommentNotification)
.outerjoin(UserCommentNotification,
UserCommentNotification.notification_id==CommentNotification.id)
.filter(UserCommentNotification.user_id==self.id))
if unconfirmed_only:
requests.filter(UserFriendshipRequestNotification.is_confirmed==False)
confirmations.filter(UserFriendshipConfirmationNotification.is_confirmed==False)
comments.filter(UserCommentNotification.is_confirmed==False)
return requests.union(confirmations, comments)
使用:user._get_notifications_query(unconfirmed_only=False).all()
生成的 SQL
SELECT anon_1.friendship_request_notifications_id AS anon_1_friendship_request_notifications_id, anon_1.friendship_request_notifications_created_at AS anon_1_friendship_request_notifications_created_at, anon_1.friendship_request_notifications_requester_id AS anon_1_friendship_request_notifications_requester_id
FROM (SELECT friendship_request_notifications.id AS friendship_request_notifications_id, friendship_request_notifications.created_at AS friendship_request_notifications_created_at, friendship_request_notifications.requester_id AS friendship_request_notifications_requester_id
FROM friendship_request_notifications LEFT OUTER JOIN users_friendship_request_notifications ON users_friendship_request_notifications.notification_id = friendship_request_notifications.id
WHERE users_friendship_request_notifications.user_id = ? UNION SELECT friendship_confirmation_notifications.id AS friendship_confirmation_notifications_id, friendship_confirmation_notifications.created_at AS friendship_confirmation_notifications_created_at, friendship_confirmation_notifications.accepter_id AS friendship_confirmation_notifications_accepter_id
FROM friendship_confirmation_notifications LEFT OUTER JOIN users_friendship_confirmation_notifications ON users_friendship_confirmation_notifications.notification_id = friendship_confirmation_notifications.id
WHERE users_friendship_confirmation_notifications.user_id = ? UNION SELECT comment_notifications.id AS comment_notifications_id, comment_notifications.created_at AS comment_notifications_created_at, comment_notifications.comment_id AS comment_notifications_comment_id
FROM comment_notifications LEFT OUTER JOIN users_comment_notifications ON users_comment_notifications.notification_id = comment_notifications.id
WHERE users_comment_notifications.user_id = ?) AS anon_1
我期待这些方面的东西
SELECT * FROM friendship_request_notifications
UNION
SELECT * FROM friendship_confirmation_notifications
UNION
SELECT * FROM comment_notifications
另外,有没有办法对 SQLAlchemy 中的聚合联合结果进行排序?
编辑
我应该提到sqlalchemy.sql.union()
产生正确的 SQL,但我不知道如何从 ORM 中使用它(返回/计算记录)。