牛客987852806号:SQL,1.连续问题可以用,subdate(day,row_number()over(partition by user_id order by day )),user_id分组求出。
2.用datediff(day,lead(day,1)over(partition by user_id order by day) )=1,进行判断,符合等0,不符合等于1,然后,得到结果sum(结果)over(partition by user_id order by day desc) ,连续的user_id,和这个值是一致的,group by 在求出来就行。小白浅见,请大佬多多见谅。