利用上一题中的代码

SELECT
    id,
    t1.job,
    score,
    t_rank
FROM(
    SELECT *,
        DENSE_RANK() OVER(PARTITION BY job ORDER BY score DESC) AS t_rank
    FROM grade
)t1
JOIN(
    SELECT 
    job,
    FLOOR((COUNT(*) + 1) / 2) AS median1,
    FLOOR((COUNT(*) + 2) / 2) AS median2
    FROM grade
    GROUP BY job
)t2
    ON t1.job = t2.job
WHERE median1 = t_rank OR median2 = t_rank
ORDER BY id

使用窗口函数求得score的排序，利用上一题中求中位数位置的方法，筛选出排序等于中位数位置的记录。