mysql-按人选前4个成绩,但至少需要两个位置

我有诸如

eventId locationId score athlete
8739    73          48  matt
8734    73          46  matt
8788    73          45  matt
8738    73          44  matt
8787    73          44  matt
8735    73          43  matt
8789    6           43  matt

我需要按人捕获前4个得分,但是前4个得分中至少有1个来自与其他3个不同的locationId

在这种情况下,我希望这个返回

eventId locationId score athlete
8739    73          48  matt
8734    73          46  matt
8788    73          45  matt
8789    6           43  matt

我试过写出将使用GROUP BY HAVING MIN(locationId)！= MAX(locationId)的查询,但是我不确定在完成ORDER BY和LIMIT的同时如何实现.

我也尝试过进行自我联接,但是我不确定如何根据s.score和score2返回最佳结果.

似乎步入正轨的自我联接的开始

SELECT s.eventid, s.locationid, athlete, score
, s2.eventid, s2.locationid, s2.athlete, score score2
FROM singles s
  INNER JOIN singles s2 ON s.athlete = s2.athlete AND s.locationid != s2.locationid
WHERE s.athlete = 'matt'
ORDER BY score DESC;

解决方法:

因此,您真正想要的是前三名,然后是第一名,这至少保证了两个位置.

这是一个相当困难的条件,但是我认为这可以解决问题：

with s as (
      select t.*,
             row_number() over (partition by athlete order by score desc) as seqnum
      from t
     ),
     s3 as (
      select s.*
      from s
      where seqnum <= 3
     )
select *
from s3
union all
(select s.*
 from s
 where ( (select count(distinct locationid) from s3) > 1 and seqnum = 4 ) or
       ( (select count(distinct locationid) from s3) = 1 and
         seqnum = (select min(seqnum)
                   from s
                   where locationid not in (select locationid from s3)
                  )
       )
);

Here是db小提琴.

标签：mysql-8-0,sql,mysql
来源： https://codeday.me/bug/20191108/2007295.html