Question

我在构建生成以下原始SQL查询所需的SQLAlchemy代码时遇到问题。

WITH RECURSIVE recruiters AS (
    SELECT 
        recruiter.id
    FROM 
        recruiter
    JOIN 
        recruiter_member ON recruiter.id = recruiter_member.recruiter_id
    WHERE
        recruiter_member.user_id = 'f12c617a-415c-4f8c-add0-81a597545be8'
    UNION ALL
    SELECT 
        children.id
    FROM 
        recrutiers AS parents, 
        recruiter AS children
    WHERE 
        children.recruiter_id = parents.id
)
SELECT
    *
FROM
    recruiters

此处的模型为Recruiter和RecruiterMember。我似乎无法让UNION正确。

Answer 1

没有更多细节，这是我能想到的最好的：

from sqlalchemy import orm

parent = orm.aliased(Recruiter)
child = orm.aliased(Recruiter)
top_q = (
    orm.query.Query([Recruiter.id.label('id')])
    .join(RecruiterMember, Recruiter.id == RecruiterMember.recruiter_id)
    .filter(RecruiterMember.user_id == 'f12c617a-415c-4f8c-add0-81a597545be8')
    .cte(recursive=True))
bottom_q = (
    orm.query.Query([child.id.label('id')])
    .join(parent, parent.id == child.recruiter_id))

final_query = top_q.union_all(bottom_q)

orm.query.Query([final_query.c.id]).with_session(session).all()

递归CTE查询

1 个答案: