概述
1，windows子句是对窗口的结果做更细粒度的划分
2、windows子句中有两种方式
rows ：按照相邻的几行进行开窗
range：按照某个值的范围进行开窗
使用方式
(rows | range) between (UNBOUNDED | [num]) PRECEDING AND ([num] PRECEDING | CURRENT ROW | (UNBOUNDED | [num]) FOLLOWING)
(rows | range) between current row AND (CURRENT ROW | (UNBOUNDED | [num]) FOLLOWING)
(rows | range) between [num] FOLLOWING AND (UNBOUNDED | [num]) FOLLOWING
1 PRECEDING 表示前一个窗口
1 FOLLOWING 表示后一个窗口
current row 表示当前窗口
UNBOUNDED：起点，
UNBOUNDED PRECEDING：表示从前面的起点，
UNBOUNDED FOLLOWING 表示到后面的终点
简单使用
姓名,购买日期,购买数量
saml,2018-01-01,10
saml,2018-01-08,55
tony,2018-01-07,50
saml,2018-01-05,46
tony,2018-01-04,29
tony,2018-01-02,15
saml,2018-02-03,23
mart,2018-04-13,94
saml,2018-04-06,42
mart,2018-04-11,75
mart,2018-04-09,68
mart,2018-04-08,62
neil,2018-05-10,12
neil,2018-06-12,80
create table sample(
name string,
dt string,
num int
)
row format delimited
fields terminated by ",";
load data local inpath '/home/homedata/sample.txt' into table sample;
rows
问题：找出最近三次的购买的数量之和（也就是这次和上两次）
select *,sum(num) over (partition by name order by dt rows between 2 PRECEDING and current row ) sum from sample ;
结果展示
range
注：1、在hive中range是不支持INTERVAL关键字的使用
2、时间字段需要转为秒值
3、值必须是计算好的 （不能是6*3600这种，需要是21600）
问题：获取每个用户最近3天购买数量
select *,sum(num) over (partition by name order by unix_timestamp(dt,"yyyy-MM-dd") range between 259200 PRECEDING and current row ) sum from sample ;
案例
rows
id dt
1 2024-04-25
1 2024-04-26
1 2024-04-27
1 2024-04-28
1 2024-04-30
1 2024-05-01
1 2024-05-02
1 2024-05-04
1 2024-05-05
2 2024-04-25
2 2024-04-28
2 2024-05-02
2 2024-05-03
2 2024-05-04
create table sql2_20(
id int,
dt string
)
row format serde 'org.apache.hadoop.hive.serde2.RegexSerDe'
with serdeproperties(
'input.regex'='(\\d+)\\s+(.+?)',
'output.format.string'='%1$s %2$s'
);
load data local inpath '/home/homedata/sql2/sql2_20.txt' into table sql2_20;
问题：现有用户登录记录表，请查询出用户连续三天登录的所有数据记录
期望结果
答案：
with t1 as (
select *,date_sub(dt,row_number() over (partition by id order by dt)) p from sql2_20
), t2 as (
select id,dt,count(*) over (partition by id,p order by dt rows between 2 PRECEDING and current row) days from t1
),t3 as (
select id,concat(date_sub(dt,2),",",date_sub(dt,1),",",dt) dts from t2 where days = 3
)
select id,dt from t3 lateral view explode(split(dts,",")) d as dt;
range
create table sql1_21(
order_id int,
user_id string,
order_status string,
operate_time string
)
row format serde 'org.apache.hadoop.hive.serde2.RegexSerDe'
with serdeproperties(
'input.regex'='(\\d+)\\s+(.+?)\\s+(.+?)\\s+(.+?)'
);
load data local inpath '/home/homedata/sql_1/sql1_21.txt' into table sql1_21;
order_id user_id order_status operate_time
1101 a 已支付 2023-01-01 10:00:00
1102 a 已取消 2023-01-01 10:10:00
1103 a 待支付 2023-01-01 10:20:00
1104 b 已取消 2023-01-01 10:30:00
1105 a 待确认 2023-01-01 10:50:00
1106 a 已取消 2023-01-01 11:00:00
1107 b 已取消 2023-01-01 11:40:00
1108 b 已取消 2023-01-01 11:50:00
1109 b 已支付 2023-01-01 12:00:00
1110 b 已取消 2023-01-01 12:11:00
1111 c 已取消 2023-01-01 12:20:00
1112 c 已取消 2023-01-01 12:30:00
1113 c 已取消 2023-01-01 12:55:00
1114 c 已取消 2023-01-01 13:00:00
问题：找出恶意购买的用户------同一个用户，在任意半小时内（含），取消订单次数>=3次的就被视为恶意买家。
结果：
with t1 as (
select order_id, user_id, unix_timestamp(operate_time) operate_time
from sql1_21 where order_status = "已取消"
)
select user_id,
count(*) over (partition by user_id order by operate_time range between 1800 preceding and current row )
from t1;