postgresql - 根据 1 个单个日期列计算一组行的 from_date 和 to_date

Question

我有一个 customer_audit 表，记录发生在客户表上的 INSERT 和 UPDATE。

客户审核

id	操作	时间戳	客户 ID	地址1	地址2
1	我	2024-10-05	100	链接st	1号
2	你	2024-10-06	100	链接st	2号
3	你	2024-10-07	100	连接路	2号
4	你	2024-10-08	100	连接路	数量200
5	我	2024-10-06	200	公园街	20号
6	你	2024-10-08	200	公园街	数量200

预期的sql输出是，对于我们感兴趣的列，在本例中是address1和address2，分别显示address1和address2的所有历史值，并根据时间戳列计算出from_date和to_date。

例如对于 customer_id 100，

客户 ID	列名称	列值	起始日期	截止日期
100	地址1	链接st	2024-10-05	2024-10-07
100	地址1	连接路	2024-10-07	null（表示列值是当前值）
100	地址2	1号	2024-10-05	2024-10-06
100	地址2	2号	2024-10-06	2024-10-08
100	地址2	数量200	2024-10-08	空

---

我通过稍微调整 mcwolf 的查询来解决这个问题，通过这样做

    select customer_id, name, 'address1' as attribute_name, address1 as "attribute_value", min(from_date) as "valid_from", max(to_date) as "valid_to"
from (
select customer_id, 
       name, 
       address1,
       "mod_timestamp" from_date,
         COALESCE(lead("mod_timestamp") over change_window, to_date('99991231', 'yyyymmdd')) to_date,
       ROW_NUMBER () OVER ( PARTITION BY customer_id, address1 ORDER BY mod_timestamp) as rownum
from customer_audit
WINDOW change_window as (PARTITION BY customer_id order by mod_timestamp)   
order by customer_id, mod_timestamp ) a
group by customer_id, name, attribute_name, "attribute_value"
order by customer_id, "valid_from";

Answer 1

...这是另一种方法...（代码中的注释）

WITH
  dates AS    -- track changes of addresses ( ..._rn = 1 )
    ( Select ca.id, customer_id, "timestamp" as ts,
             address1,
             Row_Number() Over(Partition By customer_id, address1 Order By id) addr1_rn,
             address2,
             Row_Number() Over(Partition By customer_id, address2 Order By id) addr2_rn
      From   customer_audit ca
      Order By customer_id, id
    ), 
  grid AS    -- get changes of addr1 Union All changes of addr2
    ( SELECT   f.id, f.customer_id, 
              'address1' as column_name, f.address1 as column_value, f.ts
      FROM  ( Select  * From  dates Where addr1_rn = 1 ) f
     UNION ALL 
      SELECT  f.id, f.customer_id, 
              'address2', f.address2, f.ts
      FROM ( Select  * From  dates Where addr2_rn = 1 ) f 
    )
--      M a i n    S Q L : 
Select     customer_id, column_name, column_value, 
           ts as from_date, Lead(ts) Over(Partition By customer_id, column_name
                                          Order By id) as to_date
From grid
Order By  customer_id, column_name, id

/*     R e s u l t : 
customer_id column_name column_value    from_date               to_date
----------- ----------- --------------  ----------------------  ----------------------
        100 address1    link st         2024-10-05 00:00:00     2024-10-07 00:00:00
        100 address1    link road       2024-10-07 00:00:00     null
        100 address2    number 1        2024-10-05 00:00:00     2024-10-06 00:00:00
        100 address2    number 2        2024-10-06 00:00:00     2024-10-08 00:00:00
        100 address2    number 200      2024-10-08 00:00:00     null
--
        200 address1    park st         2024-10-06 00:00:00     null
        200 address2    number 20       2024-10-06 00:00:00     2024-10-08 00:00:00
        200 address2    number 200      2024-10-08 00:00:00     null                    */