解法2里为什么要加入select distinct device_id, date from question_practice_detail这个表， 我在MySQL里 select distinct device_id, date as date1, lead(date) over (partition by device_id order by date) as date2 from ( select distinct device_id, date from question_practice_detail ) as uniq_id_date 和 select distinct device_id, date as date1, -> lead(date,1,0) over (partition by device_id order by date)as date2 -> from question_practice_detail; 得出来的表格是一样的，不影响计算平均值吧 但是为什么我提交显示是错误呢