最近有个需求，计算用户连续登录的最大天数(这里使用prestoSql，使用hive也可以)，先看下登录日志数据表hive.traffic.access_user只有两个字段：uid，day;日期辅助表hive.ods.dim_date,这个表只有一个字段day；
先说下思路，
| uid | day | rownumber | day-rownumber【days】 |
|---|---|---|---|
| 101 | 20190911 | 1 | 20190911-1=20190910 |
| 101 | 20190912 | 2 | 20190912-2=20190910 |
| 101 | 20190913 | 3 | 20190913-3=20190910 |
| 101 | 20190916 | 4 | 20190916-4=20190912 |
| 101 | 20190917 | 5 | 20190917-5=20190912 |
从上可以看到，只要是连续登录的话，day-rownumber的差值是一样的，那问题来了，这样的减法在跨月或者跨年的时候会出问题，所以我们首先将日期转换成有序的数字
select day,ROW_NUMBER() OVER(ORDER BY day) daynum from hive.ods.dim_date
接下来，我们需要将用户登录日志按照uid分组，然后按照日期排序，然后计算出rownumber
with a as (select uid,day from hive.traffic.access_user where day>=20190801 and uid<>'')
select uid,day,ROW_NUMBER() OVER(PARTITION BY uid ORDER BY uid,day) rownum from a group by day,uid
接下来就是计算差值，差值相同的代表连续登录日期，完整sql如下
with a as (select uid,day from hive.traffic.access_user where day>=20190801 and uid<>''),
b as (select uid,day,ROW_NUMBER() OVER(PARTITION BY uid ORDER BY uid,day) rownum from a group by day,uid ),
c as(select day,ROW_NUMBER() OVER(ORDER BY day) daynum from hive.ods.dim_date),
d as (select uid,b.day,daynum,rownum,daynum-rownum days from b join c on b.day=c.day )
select uid,min(day)"连续登录开始日",count(*) "连续登录天数" from d group by uid,days
end
