通过在Oracle

Question

将它们分组实现了同一行值选取所有列，具有特定特征，同子数据我有下一个结构的表：

Data structure: 
| CONTRACT | CONNECTION | STATE | 
| 1  | AAA  | Y | 
| 2  | AAA  | Y | 
| 3  | BBB  | N | 
| 4  | BBB  | N | 
| 5  | BBB  | N | 
| 6  | BBB  | N | 
| 7  | AAA  | Y | 
| 8  | CCC  | Y | 
| 9  | CCC  | N | 
| 10 | AAA  | Y | 
| 11 | CCC  | N | 

我想选择所有的行/数据考虑到由CONNECTION列定义的组为AAA，BBB和CCC组，查询必须仅选择具有值'N'的STATE列的所有行的组，但始终考虑该组的所有数据。 所以餐桌上的结果将是：

Result needed: 
| CONTRACT | CONNECTION | STATE | 
| 3  | BBB  | N | 
| 4  | BBB  | N | 
| 5  | BBB  | N | 
| 6  | BBB  | N | 

我一直在寻找的功能，如解码，存在等，但就是不明白怎么解决这个问题。

任何想法，我怎么可以为此写一个查询？

Answer 1

您可以使用SUM和HAVING：

ONLINE DEMO

SELECT * 
FROM tbl t 
WHERE 
    CONNECTION IN(
     SELECT CONNECTION 
     FROM tbl 
     GROUP BY CONNECTION 
     HAVING 
      SUM(CASE WHEN STATE = 'N' THEN 1 ELSE 0 END) > 0 
      AND SUM(CASE WHEN STATE <> 'N' THEN 1 ELSE 0 END) = 0 
    ) 

的HAVING子句中的第一个条件可以确保CONNECTION至少有一个STATE = 'N'。第二个确保CONNECTION没有其他STATE s比'N'。

Answer 2

试试这个：

SELECT CONTRACT, CONNECTION, STATE 
FROM (
    SELECT CONTRACT, CONNECTION, STATE, 
     COUNT(CASE WHEN STATE <> 'N' THEN 1 END) OVER (PARTITION BY CONNECTION) AS cnt 
    FROM mytable) t 
WHERE t.cnt = 0 

Answer 3

你可以使用窗口COUNT：

WITH cte AS (
    SELECT * 
    ,COUNT(CASE WHEN STATE = 'N' THEN 1 END) OVER(PARTITION BY CONNECTION) AS num_of_n 
    ,COUNT(CASE WHEN STATE <> 'N' OR STATE IS NULL THEN 1 END) 
      OVER(PARTITION BY CONNECTION) AS num_of_non_n 
    FROM mytable 
) 
SELECT CONTRACT, CONNECTION, STATE 
FROM cte 
WHERE num_of_n > 0 AND num_of_non_n = 0; 

LiveDemo

---

或相关性高苏bquery：

SELECT * 
FROM mytable m1 
WHERE STATE = 'N' 
    AND NOT EXISTS (SELECT 1 
        FROM mytable m2 
        WHERE m1.CONNECTION = m2.CONNECTION 
        AND (STATE <> 'N' OR STATE IS NULL)); 

LiveDemo 2