Question

我的桌子上有以下记录；

ID | NN  | MBL  | IC | OTHER
---+-----+------+----+------
1  | 123 |      |    | ac
2  |     | 544  |    | dc
3  |     |      | 524| df
4  |527  |      | 124| ff
5  |123  |      |    | tr // duplicate NN of ID 1
6  |     | 544  |    | op // duplicate MBL of ID 2
7  |     |      | 124| ii // duplicate for IC ID 4

使用select查询时，我只需要单项记录，跳过第二次出现，

select 
    ID, NN, MBL, IC, OTHER 
from 
    TABLE1  // this should return only one entry of any NN, MBL and IC

如何获取此信息，我不能对多个列使用distinct，并且还需要在选择查询中显示ID和OTHER列

预期结果如下：

1  | 123 |      |    | ac
2  |     | 544  |    | dc
3  |     |      | 524| df
4  |527  |      | 124| ff

Answer 1

您可以使用分析函数ROW_NUMBER（）计算想要的每一列的排名，并仅过滤那些等级为1的行。

这里是一个例子：

WITH testdata AS (
    SELECT 1 AS ID, 123 AS NN, NULL AS MBL, NULL AS IC, 'ac' AS OTHER FROM DUAL UNION ALL
    SELECT 2, NULL, 544 , NULL, 'dc' FROM DUAL UNION ALL
    SELECT 3, NULL, NULL, 524 , 'df' FROM DUAL  UNION ALL
    SELECT 4, 527,  NULL, 124, 'ff' FROM DUAL UNION ALL
    SELECT 5, 123,  NULL, NULL,  'tr' FROM DUAL UNION ALL
    SELECT 6, NULL, 544, NULL,   'op' FROM DUAL UNION ALL
    SELECT 7, NULL, NULL      , 124, 'ii' FROM DUAL
)
SELECT *
  FROM(SELECT ID,
              NN,
              CASE WHEN NN IS NULL THEN 1 ELSE ROW_NUMBER() OVER (PARTITION BY NN ORDER BY ID) END AS NN_RANG,
              MBL,
              CASE WHEN MBL IS NULL THEN 1 ELSE ROW_NUMBER() OVER (PARTITION BY MBL ORDER BY ID) END AS MBL_RANG,
              IC,
              CASE WHEN IC IS NULL THEN 1 ELSE ROW_NUMBER() OVER (PARTITION BY IC ORDER BY ID) END AS IC_RANG,
              OTHER
         FROM testdata
      )
 WHERE NN_RANG = 1
   AND MBL_RANG = 1
   AND IC_RANG = 1
;

希望有帮助。

从oracle中的表中仅选择不同的记录

1 个答案: