Question

首先，您是否发现此代码存在问题;

 SELECT num,
        surname, 
        firstname,
        ward 
   FROM doctor, ward WHERE num != consultant;

  NUM SURNAME    FIRSTNAME  W
---------- ---------- ---------- -
  203 Black      Peter      A
  574 Bisi       Mavis      B
  461 Boyne      Steve      B
  530 Clark      Nicola     C
  405 Mizzi      Nicola     A
  501 Mount      Mavis      A
  203 Black      Peter      A

  C NAME       CONSULTANT
  - ---------- ----------
  A Surgical          203
  B Paediatric        574
  C Medical           530

我希望输出类似于;

 461 Boyne      Steve      B
 405 Mizzi      Nicola     A
 501 Mount      Mavis      A

将会显示不相等的结果，但是当我执行命令时，结果就是这样;

       NUM SURNAME    FIRSTNAME  W
---------- ---------- ---------- -
       574 Bisi       Mavis      B
       461 Boyne      Steve      B
       530 Clark      Nicola     C
       405 Mizzi      Nicola     A
       501 Mount      Mavis      A
       203 Black      Peter      A
       461 Boyne      Steve      B
       530 Clark      Nicola     C
       405 Mizzi      Nicola     A
       501 Mount      Mavis      A
       203 Black      Peter      A

       NUM SURNAME    FIRSTNAME  W
---------- ---------- ---------- -
       203 Black      Peter      A
       574 Bisi       Mavis      B
       461 Boyne      Steve      B
       405 Mizzi      Nicola     A
       501 Mount      Mavis      A
       203 Black      Peter      A

我对Oracle很新鲜，所以这可能是一个noob错误，但任何帮助都会很棒。

Answer 1

您需要左外连接：

 SELECT num,surname, firstname,ward
 FROM doctor left outer join
      ward
      on num = consultant
 WHERE num is null;

您的查询存在的问题是，您正在doctor和ward之间进行笛卡尔积（所有组合）。然后，您将选择两个值不同的行。

写上述内容的另一种方法可能更清楚：

select d.*
from doctor d
where d.num not in (select consultant from ward);

Answer 2

<强> Here is the SQLFiddel Demo

我认为实现输出的最佳方法是使用not exsits。

以下是您可以尝试的查询

 SELECT *
   FROM doctor
  WHERE not exists (select 1 
                      from ward 
                     where num = consultant)

Answer 3

您的查询正在创建笛卡尔积而正确的查询应该是

SELECT num,surname, firstname,ward FROM doctor join ward on (c = w) WHERE num != consultant;

Answer 4

在语句中添加“和w = C”以构建连接。

或使用

SELECT num,surname, firstname,ward FROM doctor join ward on (c = w) WHERE num != consultant;

不等于（！=）未按预期运行

4 个答案: