查询满足连续任意30天的全量交易的多个商户

news2025/7/19 17:00:15

需求说明：

先说表结构把，就是一张订单表存了商户号和其他相关信息，现在要查询这个订单表中以商户为主体的连续交易，也就是每天产生至少一笔订单的商户才算连续，不知道看到的小伙伴有没有什么想法和头绪，在一般的业务场景下可能不会有这个需求，但只有遇到了才想知道解决方案当时这个需求背景是为应付某某检查所以才有这个需求，步入正题把具体分析都写在sql中不懂可以看看总结的难点，大家有场景可以试一试，一定是有多种写法的，这里我写出了两种，

最优写法:

-- with:创建临时查询命名的关键字
WITH consecutive_numbers (create_time,cust_no,group_id) AS (
  SELECT
  create_time,
  cust_no,
  -- 根据客户号分区再使用行号判断这段时间是否为连续数字,如果group_id是相同的说明就是连续的
  create_time - ROW_NUMBER() OVER (PARTITION BY cust_no ORDER BY create_time) AS group_id
  FROM (
    -- 去重意义:去重是为了保证每一天每个客户号的多笔交易的笔数只有一笔,并且使用DISTINCT去重
    -- 去重位置:只能套在子查询里去重
    -- 如果在子查询里可以使用DISTINCT to_char(t.create_time, 'yyyyMMdd')去重
    -- 也可以使用group by to_char(t.create_time, 'yyyyMMdd')去重
    
    -- 如果放在外面使用DISTINCT去重就会失效因为group_id字段是通过 ROW_NUMBER()计算出来的他在每一行有唯一的值
    -- 并且在外面使用group by去重也会报错因为group by的字段需要和查询的字段保持一致但是group by不支持ROW_NUMBER() 窗口函数
    SELECT DISTINCT
    to_char(t.create_time, 'yyyyMMdd') as create_time,
    cust_no
    FROM
    xxx.order_info t
    WHERE
    -- 这里可以指定多个商户号或是不指定查所有商户
    cust_no IN ('199900002007','199900004854')
  )
)
SELECT
    MIN(create_time) AS start_create_time,
    MAX(create_time) AS end_create_time,
		cust_no,
    COUNT(*) AS count
FROM
    consecutive_numbers
-- 对group_id,cust_no都相同的进行分组,在对分组里面的数据计算合计就能得到连续的条数count
GROUP BY
    group_id,cust_no
-- 这个判断就是判断连续的天数大于多少    
HAVING
    COUNT(*) >= 4;

写法2

 select abc.* from (with t as
(select a.CUST_NO, a.createTime, b.createTime as snap2
from (select DISTINCT cust_no, to_char(CREATE_TIME,'yyyyMMdd') as createTime from xxx.order_info where order_status='00') a,
(select DISTINCT cust_no, to_char(CREATE_TIME,'yyyyMMdd') as createTime from xxx.order_info where order_status='00') b
where a.CUST_NO = b.CUST_NO(+)
and a.createTime - 1 = b.createTime(+)
order by CUST_NO, createTime),
t1 as
(select t.*,
case
when snap2 is null then
1
else
0
end as lxzt
from t
order by CUST_NO, createTime),
t2 as
(select t1.*, sum(lxzt) over(order by CUST_NO, createTime) as fzyj
from t1
order by CUST_NO, createTime)
select CUST_NO,
--        fzyj,
((max(createTime)-min(createTime)) + 1) as 连续天数,
min(createTime) as start_snap,
max(createTime) as end_snap
from t2
where CUST_NO = '199900002007'
group by CUST_NO, fzyj
order by start_snap desc) abc;