通过代码

SELECT DISTINCT
    exam_id,
    DATE_FORMAT(start_time,'%Y%m') start_month,
    count(start_time) over(
        partition by exam_id,DATE_FORMAT(start_time,'%Y%m')
    ) month_cnt,
    count(start_time) over(
        partition by exam_id
        order by DATE_FORMAT(start_time,'%Y%m') 
    ) cum_exam_cnt
FROM
    exam_record

emm，很简单的一道题啊(连我这个笨比的代码也这么短)，所以就啥也不说了