select substring(date,9,2) day,count(question_id) question_cnt from question_practice_detail where substring(date,6,2)='08' group by day substring(date,9,2)提取出date从第九位开始的两位作为day; 筛选条件substring(date,6,2)='08',从date第6位开始提取2位并指定为08。