我将使用一个子查询：
select student_id, student_name, avg_grade, rank() over (order by avg_grade desc)from (select s.student_id, s.student_name, avg(ce.grade) as avg_grade, rank() over (order by avg(ce.grade) desc nulls last) as seqnum, count(*) over () as cnt from students s left joincourse_enrollment ceon s.student_id = ce.student_id group by s.student_id ) as ce_avgwhere seqnum <= cnt * 0.1;
您还可以使用其他窗口功能，例如
NTILE()和
PERCENTILE_DISC()。我更喜欢直接计算，因为它可以更好地控制联系的处理方式。
