显示记录不匹配在哪里条款

Question

我需要显示不匹配where子句的记录。

实施例 - select * from citytable where city in ('aa','bb','cc', 'dd','ee');

仅aa，bb，cc存在于表格中，dd & ee不存在于表格中。不过，我仍然需要显示dd & ee。

Answer 1

您可能正在寻找这样的事情。 IN条件与包含IN列表中唯一（不同）值的表的内部连接相同。你想要的是一个外连接。您需要有一个表格而不是IN列表。在下面的解决方案中，我将展示如何在运行中创建此“帮助程序”表;还有其他几种方法，这只是表明了这个想法。

select deptno, ename from emp where deptno in (10, 50, 80); 

DEPTNO ENAME 
------ ------ 
10  CLARK 
10  KING 
10  MILLER 


with h (deptno) as (
    select 10 from dual union all 
    select 50 from dual union all 
    select 80 from dual 
) 
select h.deptno, e.ename 
from h left outer join emp e 
on h.deptno = e.deptno 
; 

DEPTNO ENAME 
------ ------ 
10  CLARK 
10  KING 
10  MILLER 
50 
80 

Answer 2

我不确定你想要输出的样子。如果没有城市='ee'的数据，你想要展示什么？像这样？

SELECT * FROM 
(SELECT key AS A_key, one AS A_one, two AS A_two FROM cityTable WHERE one='aa') AS A 
JOIN 
(SELECT key AS E_key, one AS E_one, two AS E_two FROM cityTable WHERE one='ee') AS E 
ON A_key=E_key 
...etc. 

编辑：也许这是它：

Answer 3

当我明白你说的'dd' AMD 'ee'不存在于表，但你仍然需要它，就可以使用union all实现它。但请记住，“DD”和“EE”行的列将始终为空，因为没有存在的记录在你的citytable

SELECT ct.col1 AS city, ct.col2.....<all columns of your table> 
    from citytable ct 
    where city in ('aa','bb','cc') 
    UNION ALL 
    select 'dd' as city,null ,null.....<nulls as many times as the number of columns of your table> 
    from citytable ct1 
    UNION ALL 
    select 'ee' as city,null ,null.....<nulls as many times as the number of columns of your table> 
    from citytable ct2