同一张表中的行之间的差异（Oracle SQL）

Posted 2023-03-27

【中文标题】同一张表中的行之间的差异（Oracle SQL）

【英文标题】：Difference between the rows in the same table (Oracle SQL)

【发布时间】：2019-07-12 08:33:41

【问题描述】：

我真的很纠结在 Oracle SQL 中比较同一个表中的行。我需要获取之前/之后样式的所有更改。

我有这样的表：

id      date        name    action
1       01-01-2011  Alex    smth
1       05-01-2011  Alexx   smth
1       07-01-2011  Alexa   smth2
2       02-01-2012  Leo     smth3
2       05-01-2012  Leon    smth3

我需要得到这个：

id      date        field   before  after
1       05-01-2011  name    Alex    Alexx
1       07-01-2011  name    Alexx   Alexa
1       07-01-2011  action  smth    smth2
2       05-01-2012  name    Leo     Leon

我尝试将表与自身进行内部连接。我找到的方法（请在此处找到）应该可以帮助我将行与下一个内部连接返回“无效数字”错误。

Mb 有更简单的方法来完成这项任务吗？ 你能帮帮我吗？

select t1.id from tablename t1 
inner join tablename t3 on t1.id = t3.id + 1

【问题讨论】：

您能解释一下为什么结果中只有这一行：field=action：action smth smth2，但：smth 和smth3 缺失？我无法理解逻辑。

因为我只需要选择已更改的字段。

【参考方案1】：

你可以使用lag()解析函数：

with t( id, "date", name, action ) as
(
 select 1, date'2011-01-01','Alex','smth' from dual union all
 select 1, date'2011-01-05','Alexx','smth' from dual union all
 select 1, date'2011-01-07','Alexa','smth2' from dual union all
 select 2, date'2012-01-02','Leo','smth3' from dual union all
 select 2, date'2012-01-05','Leon','smth3' from dual 
), t2 as
(
select t.*, 
       lag(name,1,null) over (partition by id order by id, "date") as lg_name,
       lag(action,1,null) over (partition by id order by id, "date") as lg_action
  from t
), t3 as
(
select id, "date", 'name' as field, lg_name as before, name as after
  from t2 where name != lg_name
union all
select id, "date", 'action', lg_action, action 
  from t2 where action != lg_action
)
select * from t3 order by id, "date";

ID  date        FIELD   BEFORE  AFTER
--  ---------   -----   ------  ------
1   05-JAN-11   name    Alex    Alexx
1   07-JAN-11   action  smth    smth2
1   07-JAN-11   name    Alexx   Alexa
2   05-JAN-12   name    Leo     Leon

Demo

【讨论】：

非常感谢！这是我最需要的东西，因为我的表中有更多字段。

【参考方案2】：

我首先根据日期（每个 id）为每一行分配一个序数，然后将每一行与其下一行自我连接起来。一旦你有了它，你可以使用几个case 语句来生成前后数据：

WITH cte AS (
    SELECT id, date, name, action, 
           ROW_NUMBER() OVER (PARTITION BY id ORDER BY date ASC) AS rn
    FROM   mytable
)
SELECT a.id,
       a.date
       CASE WHEN b.name != a.name THEN 'name' ELSE 'action' END AS field,
       CASE WHEN b.name != a.name THEN b.name ELSE b.action END AS before,
       CASE WHEN b.name != a.name THEN a.name ELSE a.action END AS after
FROM   cte b
JOIN   cte a ON b.id = a.id AND b.rn = a.rn + 1

【讨论】：

如果同一条记录的两个字段都在更改，则此方法不起作用（对于给定的数据，此查询仅返回 3 条记录）。还缺少所有 case 语句的 END 关键字。

【参考方案3】：

LAG() 显然是正确的使用方法。但是，我会先取消透视：

select id, date, field, prev_value as before, value as after
from (select id, date, field, value,
             lag(value) over (partition by id, field order by date) as prev_value
      from ((select id, "date", 'name' as field, name as value
             from t
            ) union all
            (select id, "date", 'action' as field, action as value
             from t
            ) 
           ) t
      ) t
where prev_value <> value;

在较新的 Oracle 版本中，这可以使用横向连接来简化：

select id, date, field, prev_value as before, value as after
from (select t.id, t.date, x.field, x.value,
             lag(x.value) over (partition by t.id, x.field order by date) as prev_value
      from t cross join lateral
           (select 'name' as field, name as value from dual union all
            select 'action', action from dual
           ) x
      ) t
where prev_value <> value;

【讨论】：

非常感谢！它也非常有用。抱歉，无法匹配第二个答案 :(

【参考方案4】：

由于您有两个要比较的字段，我使用 UNION ALL 完成了如下：

WITH CTE AS (
    SELECT
        ID,
        DATE1,
        NAME1,
        ACTION,
        ROW_NUMBER() OVER(
            PARTITION BY ID
            ORDER BY
                DATE1 ASC
        ) AS RN
    FROM
        MYTABLE
)
-- 
 SELECT
      A.ID,
      B.DATE1,
      'name' AS FIELD,
      A.NAME1   AS BEFORE,
      B.NAME1   AS AFTER
  FROM
      CTE A
      JOIN CTE B ON B.ID = A.ID
                    AND B.RN = A.RN + 1
                        AND B.NAME1 != A.NAME1
  UNION ALL
  SELECT
      A.ID,
      B.DATE1,
      'action' AS FIELD,
      A.ACTION   AS BEFORE,
      B.ACTION   AS AFTER
  FROM
      CTE A
      JOIN CTE B ON B.ID = A.ID
                    AND B.RN = A.RN + 1
                        AND B.ACTION != A.ACTION
  ORDER BY
      ID,
      DATE1

输出：

db<>fiddle demo

干杯！！