db2 确定谁在任何30天内的消费超过了一定金额？

kwvwclae 于 2022-11-07 发布在 DB2

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我有一个表，其中列出了每个客户的交易，沿着交易发生的日期和花费的金额。我想做的是获得一个在任何30天内花费3000英镑或更多的所有客户的列表。
我可以得到一个名单，谁花了£ 3 k或更多的 * 过去 * 30天使用下面的代码，但我不知道如何适应这一点，以涵盖 * 任何 * 30天期间。任何帮助将不胜感激，请！

select  *
from
        (
        select      customer_id, sum(spend) as total_spend
        from        transaction_table
        where       transaction_date between (current date - 30 days) and current date
        group by    customer_id
        )
where   total_spend >=3000
;

db2

来源：https://stackoverflow.com/questions/70198875/identifying-who-spent-more-than-a-certain-amount-within-any-30-day-period

2条答案

按热度按时间

xzv2uavs1#

可以将SUM()与窗口函数和窗口框架30一起使用。例如：

select *
from (
  select t.*,
    sum(t.spent) over(
      partition by customer_id 
      order by julian_day(transaction_date)
      range between 30 preceding and current row
    ) as total_spend
  from transaction_table t
) x
where total_spend >= 3000

对于数据集：

CUSTOMER_ID  TRANSACTION_DATE  SPENT 
 ------------ ----------------- ----- 
 1            2021-10-01        2000  
 1            2021-10-15        1500  
 1            2021-12-01        1000  
 2            2021-11-01        2500

结果：

CUSTOMER_ID  TRANSACTION_DATE  SPENT  TOTAL_SPEND 
 ------------ ----------------- ------ ----------- 
 1            2021-10-15        1500   3500

请参阅db<>fiddle上的运行示例。

赞(0）回复(0）举报 2022-11-07

ufj5ltwl2#

请尝试以下操作。
其思想是计算每行过去30天的SPEND运行总和。

WITH TRANSACTION_TABLE (CUSTOMER_ID, TRANSACTION_DATE, SPEND) AS
(
VALUES
  (1, DATE ('2021-01-01'), 1000)
, (1, DATE ('2021-01-31'), 2000)
--, (1, DATE ('2021-02-01'), 2000)
)
SELECT DISTINCT CUSTOMER_ID
FROM
(
SELECT 
  CUSTOMER_ID
--, TRANSACTION_DATE, SPEND
, SUM (SPEND) OVER (PARTITION BY CUSTOMER_ID ORDER BY DAYS (TRANSACTION_DATE) RANGE BETWEEN 30 PRECEDING AND CURRENT ROW) AS SPEND_RTOTAL
FROM TRANSACTION_TABLE
)
WHERE SPEND_RTOTAL >= 3000

赞(0）回复(0）举报 2022-11-07

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db2 确定谁在任何30天内的消费超过了一定金额？

2条答案

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