Question

使用左外连接时如何排除一组值？

考虑以下问题：

SELECT i.id,
       i.location,
       area.description
  FROM incident_vw i,
       area_vw area
 WHERE i.area_code = area.code(+)
   AND i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')

执行查询，但不显示任何NULL区域代码值。

BE AWARE ：INCIDENT_VW.area_code可以且确实有NULL值。

如何在不使用PL / SQL的情况下排除给定区域代码集的情况下如何匹配NULL事件区域代码？

ANSI更新

使用等效的ANSI SQL也不起作用：

    SELECT i.id,
           i.location,
           area.description
      FROM incident_vw i
 LEFT JOIN area_vw area
        ON area.code = i.area_code
     WHERE i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')

解决方案

这有效：

SELECT i.id,
       i.location,
       area.description
  FROM incident_vw i,
       area_vw area
 WHERE i.area_code = area.code(+)
   AND (i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P') and i.area_code IS NULL)

谢谢大家！

Answer 1

似乎问题是IN正在删除所有为NULL的area_codes。

尝试一下：

    SELECT i.id,
           i.location,
           area.description
      FROM incident_vw i
 LEFT JOIN area_vw area
        ON area.code = i.area_code
     WHERE (i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')
            OR i.area_code IS NULL)

应该给出理想的结果......

Answer 2

我想我会试试这个：

SELECT i.id,
       i.location,
       area.description
  FROM (SELECT *
          FROM incident_vw
         WHERE i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')
            OR i.area_code IS NULL) i
     , area_vw area
 WHERE i.area_code = area.code (+)

或ANSI等价物：

   SELECT i.id,
          i.location,
          area.description
     FROM (SELECT *
             FROM incident_vw
            WHERE i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')
               OR i.area_code IS NULL) i
LEFT JOIN area_vw area
       ON i.area_code = area.code

Answer 3

我不是100％肯定，但也许你正在加入area.code上的l.area_code 当你应该加入l.area.code上的area_code时。

SELECT i.id,
       i.location,
       area.description
  FROM incident_vw i,
       area_vw area
 WHERE 1=1
   AND i.area_code = area.code(+)
   AND i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')

应该是：

SELECT i.id,
       i.location,
       area.description
  FROM incident_vw i,
       area_vw area
 WHERE 1=1
   AND i.area_code(+) = area.code
   AND i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')

Answer 4

你可以尝试一下：

   SELECT i.id,
           i.location,
           area.description
      FROM incident_vw i
 LEFT JOIN area_vw area
        ON area.code = i.area_code
     WHERE NVL(i.area_code,'something') NOT IN ('T20', 'V20B', 'V20O', 'V20P')

Answer 5

SELECT i.id,
       i.location,
       area.description
  FROM incident_vw i
  LEFT JOIN area_vw area
    ON i.area_code = area.code
  WHERE i.area_code NOT IN ('T20', 'V20B', 'V20O', 'V20P')
UNION
SELECT i.id,
       i.location,
       NULL as description
   FROM incident_vw i
   WHERE i.area_code IS NULL

Answer 6

语法看起来正确。您应该测试您的数据以确保您的理智。试试这个：

SELECT count(*) 
FROM  incident_vw i
WHERE i.area_code NOT IN 
      (SELECT a.area_code 
       FROM area_vw a);

这将告诉您是否有任何事件没有区域表中的区域。

Answer 7

我不确定我是否理解这个问题，但如果您想在排除列表中获取区域代码的NULL，那么只需将您的条件从WHERE移动到JOIN条件：

    SELECT i.id,
           i.location,
           area.description
      FROM incident_vw i
 LEFT JOIN area_vw area
        ON area.code = i.area_code
       AND area.code NOT IN ('T20', 'V20B', 'V20O', 'V20P')

从左外连接中排除设置

7 个答案: