电商项目常见连续登录，消费，日期等问题

一.数据，建表语句

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


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

二.计算逻辑

1.按照用户和日期进行分组，求每用户每天的消费

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

2.根据用户id进行开窗按照日期进行排序，如果日期减去序号数是相同的说明这几行日期是连续的

对日期进行开窗排序
select t1.id as id,t1.datestr as datestr,t1.sum_amount as 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;

上表起名为tt1，注意用到日期函数date_sub(tt1.datestr,tt1.rn)

日期减去序号再得出一列
select tt1.id,tt1.datestr,tt1.sum_amount,date_sub(tt1.datestr,tt1.rn) as grep from (select t1.id as id,t1.datestr as datestr,t1.sum_amount as 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;

临时结果

1       2019-02-08      12461.55        2019-02-07
1       2019-02-09      1052.99 2019-02-07
1       2019-02-10      85.69   2019-02-07
1       2019-02-12      769.85  2019-02-08
1       2019-02-13      943.86  2019-02-08
1       2019-02-14      538.42  2019-02-08
1       2019-02-15      369.76  2019-02-08
1       2019-02-16      369.76  2019-02-08
1       2019-02-18      795.15  2019-02-09
1       2019-02-19      715.65  2019-02-09
1       2019-02-21      537.71  2019-02-10
2       2019-02-08      12461.55        2019-02-07
2       2019-02-09      1052.99 2019-02-07
2       2019-02-10      85.69   2019-02-07
2       2019-02-12      769.85  2019-02-08
2       2019-02-13      943.86  2019-02-08
2       2019-02-14      943.18  2019-02-08
2       2019-02-15      369.76  2019-02-08
2       2019-02-18      795.15  2019-02-10
2       2019-02-19      715.65  2019-02-10
2       2019-02-21      537.71  2019-02-11
3       2019-02-08      12461.55        2019-02-07
3       2019-02-09      1052.99 2019-02-07
3       2019-02-10      85.69   2019-02-07
3       2019-02-12      769.85  2019-02-08
3       2019-02-13      943.86  2019-02-08
3       2019-02-14      276.81  2019-02-08
3       2019-02-15      369.76  2019-02-08
3       2019-02-16      369.76  2019-02-08
3       2019-02-18      795.15  2019-02-09
3       2019-02-19      715.65  2019-02-09
3       2019-02-21      537.71  2019-02-10

3.统计用户连续交易总额，连续登录天数连续登录开始时间个结束时间以及间隔天数

连续交易总额:
以id和grep进行分组，gerp相同的表示是连续交易的
select ttt1.id,ttt1.grep,sum(ttt1.sum_amount) from
(select tt1.id as id,tt1.datestr as datestr,tt1.sum_amount as sum_amount,date_sub(tt1.datestr,tt1.rn) as grep from (select t1.id as id,t1.datestr as datestr,t1.sum_amount as 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 id,grep;

临时结果：(小数列可以用round函数处理一下：round(sum(ttt1.sum_amount)))
1       2019-02-07      13600.23
1       2019-02-08      2991.6500000000005
1       2019-02-09      1510.8
1       2019-02-10      537.71
2       2019-02-07      13600.23
2       2019-02-08      3026.6499999999996
2       2019-02-10      1510.8
2       2019-02-11      537.71
3       2019-02-07      13600.23
3       2019-02-08      2730.04
3       2019-02-09      1510.8
3       2019-02-10      537.71

根据id,grep分组后查询列增加count(1)即可得到连续登录天数（代码省略）

根据id,grep分组后的datestr列，最小的即为连续登录的开始时间，最大的即为连续登录的结束时间

多出连续登录的开始时间和结束时间两列：
select ttt1.id,ttt1.grep,sum(ttt1.sum_amount),min(datestr),max(datestr),count(1) from (select tt1.id as id,tt1.datestr as datestr,tt1.sum_amount as sum_amount,date_sub(tt1.datestr,tt1.rn) as grep from (select t1.id as id,t1.datestr as datestr,t1.sum_amount as 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 id,grep;

4.间隔时间：指的是每一次连续登录时间的间隔
根据id进行聚合开窗，grep相减得到的值就是所求

举例：排序开窗后的结果
datestr    rn    grep    
2.8        1      2.7
2.9        2      2.7      2.9-2.7=2  间隔两天
2.10       3      2.7
2.13       4      2.9     2.10与2.13之间也是间隔两天
2.14       5      2.9
2.15       6      2.9
grep相同表示是连续登录的，根据grep分组后，相减得到的就是连续登录的间隔时间

聚合开窗函数：lag(字段，n):获取当前行的前几行的那一行数据
select ttt1.id,ttt1.grep,sum(ttt1.sum_amount),min(datestr),max(datestr),count(1),datediff(grep,lag(grep,1) over(partition by ttt1.id order by grep)) from (select tt1.id as id,tt1.datestr as datestr,tt1.sum_amount as sum_amount,date_sub(tt1.datestr,tt1.rn) as grep from (select t1.id as id,t1.datestr as datestr,t1.sum_amount as 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 id,grep;