下面以查询每门课程分数最高的学生以及成绩为例，演示如何查询 top N记录。下图是测试数据，表结构和相关 insert 脚本见《常用SQL之日期格式化和查询重复数据》。
select a.name,a.course,a.score from test1 a,
(select course,max(score) score from test1 group by course) b
WHERE a.course=b.course and a.score=b.score;
执行后，结果集如下：
select name,course,score from test1 a
where a.score=(select max(score) from test1 where a.course=test1.course);
或者
select name,course,score from test1 a
where not exists(select 1 from test1 where a.course=course and a.score < score);
或者
select a.* from test1 a
where 1>(select count(*) from test1 where course=a.course and score>a.score);
结果集同上图。需要注意的是如果最高分有多条，会全部查出！
以N=2为例，演示如何查询TOP N（N>1）。
如果结果集比较小，可以用程序查询单个分组结果后拼凑，也可以使用 union all。
(select name,course,score from test1 where course=‘语文‘ order by score desc limit 2)
union all
(select name,course,score from test1 where course=‘数学‘ order by score desc limit 2)
union all
(select name,course,score from test1 where course=‘英语‘ order by score desc limit 2);
select a.name,a.course,a.score
from test1 a left join test1 b on a.course=b.course and a.score<b.score
group by a.name,a.course,a.score
having count(b.id)<2
order by a.course,a.score desc;
两个表做连接查询，笛卡尔积，然后用having count(b.id)筛选出比当前条数大的条数。
select a.* from test1 a
where 2>(select count(*) from test1 where course=a.course and score>a.score)
order by a.course,a.score desc;
思路就是判断每一条记录，比a中当前记录大的条数是否为2，如果有2条比较大，则符合。筛选出全部记录，最后按课程和学分排序。但是子查询进行了n次count(*)计算，因此性能极差。
select * from test1 a where exists(select count(*) as sum from
test1 b where b.course=a.course and b.score>a.score having sum <2)
order by a.course,a.score desc;
原文：https://www.cnblogs.com/east7/p/11587365.html