连续登陆问题

连续登陆问题

在电商、物流和银行可能经常会遇到这样的需求：统计用户连续交易的总额、连续登陆天数、连续登陆开始和结束时间、间隔天数等

数据：

注意：每个用户每天可能会有多条记录

id	datestr	  amount
1,2019-02-08,6214.23 
1,2019-02-08,6247.32 
1,2019-02-09,85.63 
1,2019-02-09,967.36 
1,2019-02-10,85.69 
1,2019-02-12,769.85 
1,2019-02-13,943.86 
1,2019-02-14,538.42
1,2019-02-15,369.76
1,2019-02-16,369.76
1,2019-02-18,795.15
1,2019-02-19,715.65
1,2019-02-21,537.71
2,2019-02-08,6214.23 
2,2019-02-08,6247.32 
2,2019-02-09,85.63 
2,2019-02-09,967.36 
2,2019-02-10,85.69 
2,2019-02-12,769.85 
2,2019-02-13,943.86 
2,2019-02-14,943.18
2,2019-02-15,369.76
2,2019-02-18,795.15
2,2019-02-19,715.65
2,2019-02-21,537.71
3,2019-02-08,6214.23 
3,2019-02-08,6247.32 
3,2019-02-09,85.63 
3,2019-02-09,967.36 
3,2019-02-10,85.69 
3,2019-02-12,769.85 
3,2019-02-13,943.86 
3,2019-02-14,276.81
3,2019-02-15,369.76
3,2019-02-16,369.76
3,2019-02-18,795.15
3,2019-02-19,715.65
3,2019-02-21,537.71

建表语句

create table deal_tb(
    id string
    ,datestr string
    ,amount string
)row format delimited fields terminated by ',';

计算逻辑

• 先按用户和日期分组求和，使每个用户每天只有一条数据

 select  id
         ,datestr
         ,sum(amount) as sum_amount
 from deal_tb
 group by id,datestr

• 根据用户ID分组按日期排序，将日期和分组序号相减得到连续登陆的开始日期，如果开始日期相同说明连续登陆

 select  tt1.id
         ,tt1.datestr
         ,tt1.sum_amount
         ,date_sub(tt1.datestr,rn) as grp
 from(
     select  t1.id
             ,t1.datestr
             ,t1.sum_amount
             ,row_number() over(partition by id order by datestr) as rn
     from(
     	select  id
                 ,datestr
                 ,sum(amount) as sum_amount
        from deal_tb
        group by id,datestr
     ) t1
 ) tt1

• 统计用户连续交易的总额、连续登陆天数、连续登陆开始和结束时间、间隔天数

select  ttt1.id
        ,ttt1.grp
        ,round(sum(ttt1.sum_amount),2) as sc_sum_amount
        ,count(1) as sc_days
        ,min(ttt1.datestr) as sc_start_date
        ,max(ttt1.datestr) as sc_end_date
        ,datediff(ttt1.grp,lag(ttt1.grp,1) over(partition by ttt1.id order by ttt1.grp)) as iv_days
from(
    select  tt1.id
            ,tt1.datestr
            ,tt1.sum_amount
            ,date_sub(tt1.datestr,rn) as grp
    from(
        select  t1.id
                ,t1.datestr
                ,t1.sum_amount
                ,row_number() over(partition by id order by datestr) as rn
        from(
            select  id
                    ,datestr
                    ,sum(amount) as sum_amount
            from deal_tb
            group by id,datestr
        ) t1
    ) tt1
) ttt1
group by ttt1.id,ttt1.grp;

• 精简版

select  t1.id
        ,t1.grp
        ,round(sum(t1.sum_amount),3) as total_amount -- 连续交易总额
        ,count(1) as total_days -- 连续登录天数
        ,min(datestr) as start_date -- 连续登录开始的时间
        ,max(datestr) as end_date -- 连续登录结束的时间
        ,datediff(t1.grp,lag(t1.grp,1) over(partition by t1.id order by t1.grp)) as interval_days -- 间隔天数
from(
    select  id
            ,datestr
            ,round(sum(amount),3) as sum_amount
            ,date_sub(datestr,row_number() over(partition by id order by datestr)) as grp
    from deal_tb
    group by id,datestr
) t1
group by t1.id,t1.grp;

• 结果

1	2019-02-07	13600.23	3	2019-02-08	2019-02-10 NULL
1	2019-02-08	2991.650	5	2019-02-12	2019-02-16	1
1	2019-02-09	1510.8		2	2019-02-18	2019-02-19	1
1	2019-02-10	537.71		1	2019-02-21	2019-02-21	1
2	2019-02-07	13600.23	3	2019-02-08	2019-02-10 NULL
2	2019-02-08	3026.649	4	2019-02-12	2019-02-15	1
2	2019-02-10	1510.8		2	2019-02-18	2019-02-19	2
2	2019-02-11	537.71		1	2019-02-21	2019-02-21	1
3	2019-02-07	13600.23	3	2019-02-08	2019-02-10 NULL
3	2019-02-08	2730.04		5	2019-02-12	2019-02-16	1
3	2019-02-09	1510.8		2	2019-02-18	2019-02-19	1
3	2019-02-10	537.71		1	2019-02-21	2019-02-21	1