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，

uiddayrownumberday-rownumber【days】10120190911120190911-1=2019091010120190912220190912-2=2019091010120190913320190913-3=2019091010120190916420190916-4=2019091210120190917520190917-5=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 alt="在这里插入图片描述" />
 end