将原始 sql 转换为 sql alchemy ORM答案

【问题标题】：Converting raw sql to sql alchemy ORM将原始 sql 转换为 sql alchemy ORM
【发布时间】：2016-07-01 11:54:08
【问题描述】：

我正在寻找将以下 sql 语句转换为 sql alchemy

select c1.id, c1.repository, c1.branch, c1.failed_step, c1.stop_time
from completed_builds as c1
where c1.stop_time = (
    select max(c2.stop_time)
    from completed_builds as c2
    where c1.branch = c2.branch
    and c1.repository = c2.repository
)
and c1.repository in ('flex', 'mob', 'tv')
and c1.branch in ('stage', 'int')
order by c1.repository, c1.branch

这会输出具有最高 stop_time 值的唯一分支 + 存储库对。

非常感谢！

【问题讨论】：

select distinct completed_builds.* from completed_builds left join (select c2.* from completed_builds as c1, completed_builds as c2 where c1.branch = c2.branch and c1.repository = c2.repository and c1.stop_time > c2.stop_time ) as NOT_LARGEST ON completed_builds.id = NOT_LARGEST.id where NOT_LARGEST.id is null 这个sql语句做同样的事情。不确定此语句是否更容易转换为 sqlalchemy？
请使用问题上的编辑链接添加更多信息。
我们无法为您编写代码。你试过什么？
多个 where 语句有问题吗？ than you might find your answer here 使用 sqlalchemy.and_

标签： sql sqlalchemy

【解决方案1】：

使用子查询添加过滤器。

subq = (db.query(CompletedBuilds.id, 
                 func.max(CompletedBuilds.stop_time).label(‘max_stop_time’)
                 CompletedBuilds.branch,
                 CompletedBuilds.repository)
       .subquery())

q = (db.query(CompletedBuilds)
    .join(subq, subq.id  == CompletedBuilds.id)
    .filter( subq.branch == CompletedBuilds.branch,
            subq.repository == CompletedBuilds.repository, 
            CompletedBuilds.repository.in_(‘flex’, ‘mob’, ‘tv’), 
            CompletedBuilds.branch.in_(‘stage’, ‘int’))
    .order_by(CompletedBuilds.repository, CompletedBuilds.branch))

【讨论】：