关于我为这篇文章SQL Group By and Limit issue接受的答案,我需要弄清楚如何使用SQLAlchemy创建该查询。作为参考,我需要运行的查询是:
SELECT t.id, t.creation_time, c.id, c.creation_time
FROM (SELECT id, creation_time
FROM thread
ORDER BY creation_time DESC
LIMIT 5
) t
LEFT OUTER JOIN comment c ON c.thread_id = t.id
WHERE 3 >= (SELECT COUNT(1)
FROM comment c2
WHERE c.thread_id = c2.thread_id
AND c.creation_time <= c2.creation_time
)
我有查询的前半部分,但我正在努力学习WHERE子句的语法以及如何将它与JOIN结合起来。有人有什么建议吗?
谢谢!
编辑:第一次尝试似乎搞乱了.filter()调用:
c = aliased(Comment)
c2 = aliased(Comment)
subq = db.session.query(Thread.id).filter_by(topic_id=122098).order_by(Thread.creation_time.desc()).limit(2).offset(2).subquery('t')
subq2 = db.session.query(func.count(1).label("count")).filter(c.id==c2.id).subquery('z')
q = db.session.query(subq.c.id, c.id).outerjoin(c, c.thread_id==subq.c.id).filter(3 >= subq2.c.count)
这将生成以下SQL:
SELECT t.id AS t_id, comment_1.id AS comment_1_id
FROM (SELECT count(1) AS count
FROM comment AS comment_1, comment AS comment_2
WHERE comment_1.id = comment_2.id) AS z, (SELECT thread.id AS id
FROM thread
WHERE thread.topic_id = :topic_id ORDER BY thread.creation_time DESC
LIMIT 2 OFFSET 2) AS t LEFT OUTER JOIN comment AS comment_1 ON comment_1.thread_id = t.id
WHERE z.count <= 3
请注意,子查询排序不正确,并且subq2以某种方式从注释中选择两次。手动修复可以得到正确的结果,我只是不确定如何让SQLAlchemy做到正确。
答案 0 :(得分:1)
试试这个:
c = db.aliased(Comment, name='c')
c2 = db.aliased(Comment, name='c2')
sq = (db.session
.query(Thread.id, Thread.creation_time)
.order_by(Thread.creation_time.desc())
.limit(5)
).subquery(name='t')
sq2 = (
db.session.query(db.func.count(1))
.select_from(c2)
.filter(c.thread_id == c2.thread_id)
.filter(c.creation_time <= c2.creation_time)
.correlate(c)
.as_scalar()
)
q = (db.session
.query(
sq.c.id, sq.c.creation_time,
c.id, c.creation_time,
)
.outerjoin(c, c.thread_id == sq.c.id)
.filter(3 >= sq2)
)