如何获取行的最后更改时间戳？

Posted 2023-03-28

【中文标题】如何获取行的最后更改时间戳？

【英文标题】：How the get last change timestamp for row?

【发布时间】：2017-04-29 12:07:17

【问题描述】：

我有一个这样的数据集（下面的 DDL）：

+----+------------------+----------------------+---------------------+
| ID | NAME             | EMAIL                | LAST_UPD            |
+----+------------------+----------------------+---------------------+
| 1  | JOHN SMITH       | [e-mail] | 29/04/2017 10:50:51 |
+----+------------------+----------------------+---------------------+
| 1  | J SMITH          | [e-mail] | 29/04/2017 10:51:15 |
+----+------------------+----------------------+---------------------+
| 1  | J SMITH          | [e-mail] | 29/04/2017 10:51:36 |
+----+------------------+----------------------+---------------------+
| 1  | JOHN JAMES SMITH | [e-mail] | 29/04/2017 10:52:11 |
+----+------------------+----------------------+---------------------+
| 2  | JAMES FORD       | [e-mail] | 29/04/2017 10:52:57 |
+----+------------------+----------------------+---------------------+
| 2  | JAMES FORD       | [e-mail]     | 29/04/2017 10:53:17 |
+----+------------------+----------------------+---------------------+
| 2  | JAMES FORD       | [e-mail]     | 29/04/2017 11:47:15 |
+----+------------------+----------------------+---------------------+

我正在尝试获取名称和电子邮件列的最后更新日期（更改的时间戳）以及 ID 中这些字段的相应值。如果给定的属性没有改变，则应该接收到最小的 LAST_UPD。我尝试了这个并得到了我想要的值，但是如何为给定的 ID“挤压”这个值？

SELECT 
ID,
NAME,
CASE 
  WHEN LAG(NAME)OVER(PARTITION BY ID ORDER BY LAST_UPD) != NAME 
  THEN LAST_UPD 
  WHEN LEAD(NAME)OVER(PARTITION BY ID ORDER BY LAST_UPD) = NAME
  THEN LAST_UPD
END NAME_CHANGED,
EMAIL, 
CASE 
  WHEN LAG(EMAIL)OVER(PARTITION BY ID ORDER BY LAST_UPD) != EMAIL 
  THEN LAST_UPD 
  WHEN LEAD(EMAIL)OVER(PARTITION BY ID ORDER BY LAST_UPD) = EMAIL
  THEN LAST_UPD
END EMAIL_CHANGED
FROM CUSTOMER
;

结果应该是这样的：

+----+------------------+---------------------+----------------------+---------------------+
| ID | NAME             | NAME_CHANGED        | EMAIL                | EMAIL_CHANGED       |
+----+------------------+---------------------+----------------------+---------------------+
| 1  | JOHN JAMES SMITH | 29/04/2017 10:52:11 | [e-mail] | 29/04/2017 10:50:51 |
+----+------------------+---------------------+----------------------+---------------------+
| 2  | JAMES FORD       | 29/04/2017 10:52:57 | [e-mail]     | 29/04/2017 10:53:17 |
+----+------------------+---------------------+----------------------+---------------------+

DDL：

CREATE TABLE CUSTOMER 
(
  ID VARCHAR2(20) 
, NAME VARCHAR2(50) 
, EMAIL VARCHAR2(50) 
, LAST_UPD DATE 
);

REM INSERTING into CUSTOMER
SET DEFINE OFF;
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('1','JOHN SMITH','[e-mail]',to_date('29/04/2017 10:50:51','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('1','J SMITH','[e-mail]',to_date('29/04/2017 10:51:15','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('1','J SMITH','[e-mail]',to_date('29/04/2017 10:51:36','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('1','JOHN JAMES SMITH','[e-mail]',to_date('29/04/2017 10:52:11','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('2','JAMES FORD','[e-mail]',to_date('29/04/2017 10:52:57','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('2','JAMES FORD','[e-mail]',to_date('29/04/2017 10:53:17','DD/MM/YYYY HH24:MI:SS'));
Insert into CUSTOMER (ID,NAME,EMAIL,LAST_UPD) values ('2','JAMES FORD','[e-mail]',to_date('29/04/2017 11:47:15','DD/MM/YYYY HH24:MI:SS'));
COMMIT;

SELECT * FROM CUSTOMER;

【参考方案1】：

我认为关键思想是从指示电子邮件或名称是否更改的标志开始。您可以使用延迟来获得此信息。而且，使用正确的逻辑，您甚至可以将第一条记录标记为更改。

然后，您需要标记为更改的每一列的最后一条记录。以下代码使用first_value() 函数执行此操作——因为它可以忽略空值：

select distinct id,
       first_value((case when name_changed then name end) ignore nulls) over (partition by id order by last_upd desc),
       max(case when name_changed then last_upd end) over (partition by id) as last_upd_name,
       first_value((case when email_changed then email end) ignore nulls) over (partition by id order by last_upd desc),
       max(case when email_changed then last_upd end) over (partition by id) as last_upd_email
from (select c.*,
             (case when c.name = lag(c.name) over (partition by c.id over order by c.last_upd) as name_changed,
                   then 0 else 1
              end) as name_changed,
             (case when c.email = lag(c.email) over (partition by c.id over order by c.last_upd) as email_change
                   then 0 else 1
              end) as email_changed
      from customer c
     ) c;

【讨论】：

谢谢，这似乎不适用于 Oracle 数据库，试图弄清楚如何更改语法。

@jrara 。 . .这是为 Oracle 设计的。这可能是一些简单的语法错误，例如不匹配的括号或缺少逗号。

【参考方案2】：

select id, name, max(nc) name_changed, email, max(mc) email_changed
  from (
   select id,
          first_value(name) over(partition by id order by last_upd desc) name,
          case lead(name) over(partition by id order by last_upd desc)
               when name then NULL else last_upd end nc,
          first_value(email) over(partition by id order by last_upd desc) email,
          case lead(email) over(partition by id order by last_upd desc)
               when email then NULL else last_upd end mc
     from CUSTOMER
  )
 group by id,name,email

【讨论】：

谢谢，这也有效。 @Gordon Linoff 提供的解决方案在我的原始数据集（5,500 万行）上稍慢。

@jrara 略微更正。 -1 窗口排序。比较速度，如果不难的话。您的数据样本中有多少结果行？

【参考方案3】：

with data as
(
  select ROWNUM AS RN, I.* 
  from 
  (
    select id,COL, VAL, LAST_UPD from customer
    unpivot(val for (col) in (NAME, EMAIL)) order by id, col, last_upd
  ) I
)
,
cte (rn, id, col, val, last_upd, dummy) as
(
  select rn, id, col, val, last_upd, 1 
  from data 
  where rn in (select rn from (select rn, min(rn) over (partition by id, col) m from data) where rn = m)
  union all
  select 
    data.rn, data.id, data.col, 
    case when cte.val = data.val then cte.val else data.val end, 
    case when cte.val = data.val then cte.last_upd else data.last_upd end, 
    cte.dummy+1 
  from 
    data, cte 
  where 
    cte.rn + 1 = data.rn and cte.col = data.col and cte.id = data.id
)
,
rs as
(
  select * from (
  select cte.*, max(dummy) over (partition by id, col) m from cte
  order by rn, id, col) where dummy = m
)
SELECT 
  n.ID, n.val as NAME, n.last_upd as NAME_CHANGED, 
  m.VAL as EMAIL, m.lAST_UPD as EMAIL_CHANGED
FROM
  (select * from rs where col = 'NAME') n 
  join
  (select * from rs where col = 'EMAIL') m
  on (n.id = m.id)
;

【参考方案4】：

基于@Gordon Linoff 的回答，针对 Oracle 数据库进行了修改，这可行：

WITH CUST AS (
SELECT ID,
      NAME,
      EMAIL,
      LAST_UPD,
      CASE WHEN NAME = LAG(NAME) OVER (PARTITION BY ID ORDER BY LAST_UPD) THEN 0 ELSE 1 END AS NAME_CHANGED,
      CASE WHEN EMAIL = LAG(EMAIL) OVER (PARTITION BY ID ORDER BY LAST_UPD) THEN 0 ELSE 1 END AS EMAIL_CHANGED
FROM CUSTOMER
)
SELECT DISTINCT CUST.ID,
       FIRST_VALUE(CASE WHEN NAME_CHANGED = 1 THEN CUST.NAME END) IGNORE NULLS OVER (PARTITION BY ID ORDER BY LAST_UPD DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS NAME,
       MAX(CASE WHEN CUST.NAME_CHANGED = 1 THEN CUST.LAST_UPD END) OVER (PARTITION BY CUST.ID) AS LAST_UPD_NAME,
       FIRST_VALUE(CASE WHEN EMAIL_CHANGED = 1 THEN EMAIL END) IGNORE NULLS OVER (PARTITION BY ID ORDER BY LAST_UPD DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS EMAIL,
       MAX(CASE WHEN CUST.EMAIL_CHANGED  = 1 THEN CUST.LAST_UPD END) OVER (PARTITION BY CUST.ID) AS LAST_UPD_EMAIL
FROM CUST
ORDER BY CUST.ID
;      

结果：

+----+------------------+---------------------+----------------------+---------------------+
| ID | NAME             | NAME_CHANGED        | EMAIL                | EMAIL_CHANGED       |
+----+------------------+---------------------+----------------------+---------------------+
| 1  | JOHN JAMES SMITH | 29/04/2017 10:52:11 | [e-mail] | 29/04/2017 10:50:51 |
+----+------------------+---------------------+----------------------+---------------------+
| 2  | JAMES FORD       | 29/04/2017 10:52:57 | [e-mail]     | 29/04/2017 10:53:17 |
+----+------------------+---------------------+----------------------+---------------------+