如何列出最大值为一列的行

Question

我有一些数据库，例如下面的例子。

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

如何以Max（N1）列出所有行，并按C1，C2分组，如下图所示？

Answer 1

问题中对要求的描述不清楚，我猜测每个C1 + c2 组的max（n1）应计算。
如果是这种情况，那么您可以通过以下方式使用MAX() OVER ()分析函数：

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 |

然后将上述查询包装为子查询，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 |

演示：http://sqlfiddle.com/#!4/d2fb9/4

Answer 2

一种方法是使用KEEP .. DENSE_RANK

SQL Fiddle

查询1 ：

    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 C1
    ,C2
    ,MAX(N1) AS N1
    ,MAX(N2) KEEP (
        DENSE_RANK FIRST ORDER BY N1 DESC
        ) AS N2
FROM TB
GROUP BY C1
    ,C2

<强> Results ：

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