У меня есть несколько разных таблиц уведомлений, и я хотел бы объединить их, чтобы показать пользователю все их уведомления. Тем не менее, профсоюз не работает как следует.

Python code

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)

Использование: user._get_notifications_query (unsfirmed_only = False) .all ()

SQL generated

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

I expect something along these lines

SELECT * FROM friendship_request_notifications
UNION
SELECT * FROM friendship_confirmation_notifications
UNION 
SELECT * FROM comment_notifications

Кроме того, есть ли способ сортировать агрегированные результаты объединения из SQLAlchemy?

EDIT

Я должен упомянуть, чтоsqlalchemy.sql.union() производит правильный SQL, но я не знаю, как использовать это из ORM (возврат / подсчет записей).