hive sql之：最大登录天数，获取连续登录指定天数

Posted 2021-08-29 to.to

create table test2(
  id string,
  pday string
);

INSERT INTO test2(id,pday) values ('A','20190701');
INSERT INTO test2(id,pday) values ('A','20190702');
INSERT INTO test2(id,pday) values ('A','20190703');
INSERT INTO test2(id,pday) values ('A','20190704');
INSERT INTO test2(id,pday) values ('A','20190706');
INSERT INTO test2(id,pday) values ('A','20190707');
INSERT INTO test2(id,pday) values ('A','20190708');
INSERT INTO test2(id,pday) values ('A','20190711');
INSERT INTO test2(id,pday) values ('A','20190712');

INSERT INTO test2(id,pday) values ('B','20190629');
INSERT INTO test2(id,pday) values ('B','20190630');
INSERT INTO test2(id,pday) values ('B','20190701');
INSERT INTO test2(id,pday) values ('B','20190704');
INSERT INTO test2(id,pday) values ('B','20190706');

最大登录天数

select 
  t2.id,
  max(t2.num)
from 
(
	select 
	t.id as id,
	count(t.sub) num
	from 
	(
		select 
			id,
			pday,
			date_sub(
				from_unixtime(unix_timestamp(pday,'yyyyMMdd'),'yyyy-MM-dd'),
				row_number() over(partition by id order by pday)
			) as sub
		from test2
	) as t
	group by t.id,t.sub
) t2
group by t2.id;

获取连续登录指定天数的：

select
t.id as id,
t.pday as pday,
date_sub(t.pday,rn) as data_sub,
t.rn as rn
from 
(
	select 
	id,
	from_unixtime(unix_timestamp(pday,'yyyyMMdd'),'yyyy-MM-dd') as pday,
	row_number() over(partition by id order by pday desc) as rn
	from test2
) t
where t.rn = 3;

===============================================
datediff的用法

select *
from 
(
select 
id,
from_unixtime(unix_timestamp(pday,'yyyyMMdd'),'yyyy-MM-dd') as pday,
date_sub(
from_unixtime(unix_timestamp(pday,'yyyyMMdd'),'yyyy-MM-dd'),
row_number() over(partition by id order by pday)
) date_sub
from test2
) t2 
where datediff(t2.pday,t2.date_sub) > 2;