Question

我真的比较比较Oracle SQL中同一表中的行。我需要获得样式前后的所有更改。

我有这样的桌子：

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

我试图将表本身与其内部连接。我发现的方法（请在这里找到）可以帮助我将内在行与下一行连接起来，返回“无效数字”错误。

Mb有执行此任务的更简单方法吗？你能帮我吗？

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

Answer 1

我首先要根据日期（按id）为每一行分配一个序数，然后将每一行与其下一行自动连接起来。一旦有了这些，就可以使用几个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

Answer 2

您可以使用lag()分析函数：

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

Answer 3

LAG()显然是正确的使用方法。但是，我将首先取消枢纽：

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Oracle的最新版本中，可以使用横向联接来简化此操作：

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;

Answer 4

由于您有两个要比较的字段，因此我使用UNION ALL进行了以下操作：

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！

同一张表中的行之间的差异（Oracle SQL）

4 个答案: