Question

我有一张这样的表

ID | Type | Val0 | Creation_date
1  |  A   |  V1  | 24sep, 2011
2  |  B   |  X1  | 15 dec, 2001
3  |  A   |  V1  | 27oct, 2008
4  |  A   |  V2  | 19oct, 2003
5  |  B   |  X2  | 15 dec, 2000
6  |  C   |  X3  | 23Dec, 2001
7  |  B   |  X2  | 15 dec, 1990

我想只保留特定类型的最新值，并提取旧的值。因此，对于上表，我希望将这些项目提取出来 - V2，X2，我可以在以后删除。

由于A的最新值为V1，因此B的最新值为X1，而C的最新值为X3。

有人可以建议我如何使用SQL来做到这一点吗？

Answer 1

如果您只想要这些值，可以使用keep：

select type,
       max(val0) keep (dense_rank first order by creation_date asc) as latest_val0
from t
group by type;

如果您想要整行，请使用row_number()：

select t.*
from (select t.*,
             row_number() over (partition by type order by creation_date asc) as seqnum
      from t
     ) t
where seqnum = 1;

Answer 2

Oracle安装程序：

CREATE TABLE table_name ( ID, Type, Val, Creation_date ) AS
SELECT 1, 'A', 'V1', DATE '2011-09-24' FROM DUAL UNION ALL
SELECT 2, 'B', 'X1', DATE '2001-12-15' FROM DUAL UNION ALL
SELECT 3, 'A', 'V1', DATE '2008-10-27' FROM DUAL UNION ALL
SELECT 4, 'A', 'V2', DATE '2003-10-19' FROM DUAL UNION ALL
SELECT 5, 'B', 'X2', DATE '2000-12-15' FROM DUAL UNION ALL
SELECT 6, 'C', 'X3', DATE '2001-12-23' FROM DUAL UNION ALL
SELECT 7, 'B', 'X2', DATE '1990-12-15' FROM DUAL;

<强>查询：

SELECT ID,
       TYPE,
       val,
       Creation_date
FROM   (
  SELECT t.*,
         LAST_VALUE( val ) OVER (
             PARTITION BY type
             ORDER BY creation_date
             ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
          ) AS last_val
  FROM   table_name t
)
WHERE val <> last_val;

<强>输出：

        ID TYPE VAL CREATION_DATE     
---------- ---- --- -------------------
         4 A    V2  2003-10-19 00:00:00 
         7 B    X2  1990-12-15 00:00:00 
         5 B    X2  2000-12-15 00:00:00

如何为每个组分配旧值

2 个答案: