HIVE窗口函数按日期累加 (取消费满50的日期，满100的日期)

取每个顾客消费满50、满100时的日期

表如下图：

name	costdate	num
jack	2015-01-01	10
tony	2015-01-02	15
jack	2015-02-03	23
tony	2015-01-04	29
jack	2015-01-05	46
jack	2015-04-06	42
tony	2015-01-07	50
jack	2015-01-08	55
mart	2015-04-08	62
mart	2015-04-09	68
neil	2015-05-10	12
mart	2015-04-11	75
neil	2015-06-12	80
mart	2015-04-13	94

 

1.计算出每天的累加和、以及累积和-50、累加和-100  （设此临时表为qq）

select 
name,
costdate,
--每天消费金额
num,
--累加和
sumnum,
sumnum-50 as num_50,
sumnum-100 as num_100
from
(
select name,costdate,num,sum(num) over(partition by name order by costdate) as sumnum
from aa
)aaa 

2.分别对 累加和-50  和  累加和-100  的差值筛选（大于等于0）并且按name分组排序 取排序后为第一的值

再用 累加和-50  排序后为第一的值  左连接  累加和-100  排序后为第一的值  （qq为步骤1的临时表）

select
mm.name,
--消费到50的日期
mm.costdate as costdate_50,
--减50
num_50,
--排序
num50,
--消费到50的金额
mm.sumnum as sumnum_50,
--消费到100的日期
nn.costdate as costdate_100,
--减100
num_100, 
--排序
num100,
--消费到100的金额
nn.sumnum as sumnum_100
from
(
select name,costdate,num,sumnum,num_50, num50
from
(select name,costdate,num,sumnum,num_50,row_number() over(partition by name order by num_50) as num50
from qq
where num_50>=0
)m
where  num50=1
)mm

left join 

(
select
name,costdate,num,sumnum,num_100, num100
from
(select name,costdate,num,sumnum,num_100,row_number() over(partition by name order by num_100) as num100
from qq
where num_100>=0
)n
where n.num100=1
)nn on mm.name=nn.name

 

mm.name	costdate_50	num_50	num50	sumnum_50	costdate_100	num_100	num100	sumnum_100
jack	2015-01-05	6	1	56	2015-01-08	11	1	111
mart	2015-04-08	12	1	62	2015-04-09	30	1	130
neil	2015-06-12	42	1	92	NULL	NULL	NULL	NULL
tony	2015-01-07	44	1	94	NULL	NULL	NULL	NULL