Question

如何在Oracle中执行以下操作：

我有一张（简化的）表格：

+-----+-----+-----+
|  a  |  b  | ... |
+-----+-----+-----+
|  1  |  7  | ... |
|  2  |  5  | ... |
|  1  |  7  | ... |
+-----+-----+-----+

a作为一个人的唯一标识符，b是我感兴趣的跨行匹配的字段。我如何构建一个基本上说＆＃34;给我个人ID的查询，其中该人有多个b值（即重复）＆＃34;？

到目前为止，我已经尝试过：

SELECT a FROM mytable GROUP BY a HAVING COUNT(DISTINCT b) > 1;

这感觉很接近，除了它只是给了我用户ID，其中用户有多个唯一 b，我怀疑它来自DISTINCT部分，但我不确定如何更改查询以实现我想要的。

Answer 1

尝试

group by a,b having count(b) > 1

您的7,5,7计为2（一7，一5）。这一个将计算任何分组中的总B数，因此您将获得1,7 - ＆gt; 2和1,5 - ＆gt; 1

Answer 2

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE mytable ( a, b ) AS
SELECT LEVEL, LEVEL FROM DUAL CONNECT BY LEVEL <= 2000
UNION ALL
SELECT LEVEL *2, LEVEL * 2 FROM DUAL CONNECT BY LEVEL <= 1000;

查询1 ：

WITH data AS (
 SELECT a
 FROM mytable
 GROUP BY a
 HAVING COUNT(b) > COUNT( DISTINCT b )
 ORDER BY a
),
numbered AS (
 SELECT a,
 ROWNUM AS rn
 FROM data
)
SELECT a
FROM numbered
WHERE rn <= 20

<强> Results ：

|  A |
|----|
|  2 |
|  4 |
|  6 |
|  8 |
| 10 |
| 12 |
| 14 |
| 16 |
| 18 |
| 20 |
| 22 |
| 24 |
| 26 |
| 28 |
| 30 |
| 32 |
| 34 |
| 36 |
| 38 |
| 40 |

Oracle为给定列选择具有相同值的两个（或更多）相邻行

2 个答案: