SELECT uid,COUNT(uid) exam_complete_cnt FROM( SELECT uid,submit_time, DENSE_RANK() OVER (PARTITION BY uid ORDER BY DATE_FORMAT(start_time,'%Y%m') DESC) month_rank FROM exam_record )t1 WHERE month_rank<=3 GROUP BY uid HAVING COUNT(uid) = COUNT(submit_time) ORDER BY exam_complete_cnt DESC,uid DESC;看uid和submit_time是否一致数量。



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