user登陆表数据如下，求出连续登录3天及以上的用户

方法1：排序，dt列求出本行和前面第2行的日期差，等于2则三天连续

SELECT DISTINCT uid
FROM
(SELECT uid,dt,lag(dt,2) over(PARTITION BY uid ORDER BY dt) AS lag_dt
FROM USER
)aa WHERE DATEDIFF(dt,lag_dt)=2

方法2：排序，dt减去排序序号的日期差相等的数量大于等于3 （推荐！！！）

SELECT uid
FROM
(SELECT uid,dt,rn,DATE_SUB(dt,INTERVAL rn DAY) AS diffFROM(SELECT uid,dt,row_number() over(PARTITION BY uid ORDER BY dt) AS rn FROM USER)aa
)bb
GROUP BY uid,diff HAVING COUNT(1)>=3