oracle - 如何列出具有一列最大值的行

Question

我有一些数据库，如下例所示。

WITH TB AS(
     SELECT 'A' C1, 'B' AS C2, 1 AS N1, 1 AS N2 FROM DUAL UNION ALL
     SELECT 'A' C1, 'B' AS C2, 2 AS N1, 2 AS N2 FROM DUAL UNION ALL
     SELECT 'A1' C1, 'B1' AS C2, 1 AS N1, 3 AS N2 FROM DUAL UNION ALL
     SELECT 'A1' C1, 'B1' AS C2, 2 AS N1, 4 AS N2 FROM DUAL UNION ALL
     SELECT 'A1' C1, 'B1' AS C2, 3 AS N1, 1 AS N2 FROM DUAL UNION ALL
     SELECT 'A2' C1, 'B2' AS C2, 1 AS N1, 6 AS N2 FROM DUAL 
)
SELECT * FROM TB

如何列出具有 Max(N1) 的所有行，并按 C1、C2 分组，如下图所示？

Answer 1

问题中的要求描述不清楚，我猜测应该为每组C1+c2计算max(n1)。
如果是这种情况，那么您可以这样使用 MAX() OVER () 解析函数:

SELECT tb.*,
       max( n1 ) over (partition by c1, c2 ) xxxx
FROM TB;

| C1 | C2 | N1 | N2 | XXXX |
|----|----|----|----|------|
|  A |  B |  1 |  1 |    2 |
|  A |  B |  2 |  2 |    2 |
| A1 | B1 |  1 |  3 |    3 |
| A1 | B1 |  2 |  4 |    3 |
| A1 | B1 |  3 |  1 |    3 |
| A2 | B2 |  1 |  6 |    1 |

然后将上面的查询包装为子查询，并过滤掉不需要的行:

SELECT c1,c2,n1,n2 FROM (
  SELECT tb.*,
       max( n1 ) over (partition by c1, c2 ) xxxx
  FROM TB
)
WHERE n1 = xxxx

| C1 | C2 | N1 | N2 |
|----|----|----|----|
|  A |  B |  2 |  2 |
| A1 | B1 |  3 |  1 |
| A2 | B2 |  1 |  6 |

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演示:http://sqlfiddle.com/#!4/d2fb9/4