SQLAlchemy: filter by membership in at least one many-to-many related table

不问归期 提交于 2019-11-28 21:18:34

OK, after a lot of research, I realized that it was my own ignorance of SQL terminology that was holding me back. My search for a solution to find users belonging to "at least one of" the list of groups should have been to find users belonging to "any" of the list of groups. The any ORM function from SQLAlchemy does exactly what I needed, like so:

session.query(ZKUser).filter(ZKUser.groups.any(ZKGroup.id.in_([1,2,3])))

That code emits this SQL (on MySQL 5.1):

SELECT * FROM users 
WHERE EXISTS (
    SELECT 1 FROM user_groups, groups 
    WHERE users.id = user_groups.contact_id 
        AND groups.id = user_groups.group_id 
        AND groups.id IN (%s, %s, %s)
    )

According to the docs for any, the query will run faster if you use an explicit join instead:

Because any() uses a correlated subquery, its performance is not nearly as good when compared against large target tables as that of using a join.

In your case, you could do something like:

users = (
    session.query(ZKUser)
    .join(user_groups)
    .filter(user_groups.columns.group_id.in_([1, 2, 3]))
)

This emits SQL like:

SELECT *
FROM users
JOIN user_groups ON users.id = user_groups.user_id 
WHERE user_groups.group_id IN (1, 2, 3)

You can use in_:

session.query(ZKUser).filter(ZKGroup.id.in_([1,2])).all()
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!