最近有個(gè)需求，計(jì)算用戶連續(xù)登錄的最大天數(shù)(這里使用prestoSql，使用hive也可以)，先看下登錄日志數(shù)據(jù)表hive.traffic.access_user只有兩個(gè)字段：uid，day;日期輔助表hive.ods.dim_date,這個(gè)表只有一個(gè)字段day；
先說(shuō)下思路，

uiddayrownumberday-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

從上可以看到，只要是連續(xù)登錄的話，day-rownumber的差值是一樣的，那問(wèn)題來(lái)了，這樣的減法在跨月或者跨年的時(shí)候會(huì)出問(wèn)題，所以我們首先將日期轉(zhuǎn)換成有序的數(shù)字

select day,ROW_NUMBER() OVER(ORDER BY day) daynum from hive.ods.dim_date

接下來(lái)，我們需要將用戶登錄日志按照uid分組，然后按照日期排序，然后計(jì)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

接下來(lái)就是計(jì)算差值，差值相同的代表連續(xù)登錄日期，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)"連續(xù)登錄開(kāi)始日",count(*) "連續(xù)登錄天數(shù)" from d group by uid,days

end

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