Sql 如何让不具备特定技能的员工脱颖而出

我有一个有两列的表。员工id，每个员工和员工技能集的下一列都是唯一的。一名员工可以拥有多个技能集。如果A、B、c、D、E是员工可以拥有的五种技能集，那么如何检索没有技能集“c”的不同员工列表

employee_id skillset
1           A
1           C
2           E
3           A
3           B
3           C
4           D
4           C
5           B

我尝试过自连接和其他方法，但不起作用

select distinct employee_id from employee_skillset where skillset not like 'C'

当我运行查询时，它仍然会给我员工id，这些id的skillset为c

您可以按员工id分组，并在HAVING子句中设置一个条件：

select employee_id 
from employee_skillset 
group by employee_id
having sum(case when skillset = 'C' then 1 else 0 end) = 0

或不存在：

select distinct s.employee_id 
from employee_skillset s
where not exists (
  select 1 from employee_skillset
  where employee_id = s.employee_id and skillset = 'C'
)  

---

您对数据集的预期结果是什么？2号和5号

为什么不像下面这样

SELECT DISTINCT employee_id
FROM Table1
WHERE skillset <> 'C';

减集运算符是一个选项：

SQL> with employee_skillset (employee_id, skillset) as
  2    (select 1, 'a' from dual union all
  3     select 1, 'c' from dual union all
  4     select 2, 'e' from dual union all
  5     select 3, 'a' from dual union all
  6     select 3, 'b' from dual union all
  7     select 3, 'c' from dual union all
  8     select 4, 'd' from dual union all
  9     select 4, 'c' from dual union all
 10     select 5, 'b' from dual
 11    )
 12  select employee_id from employee_skillset
 13  minus
 14  select employee_id from employee_skillset where skillset = 'c';

EMPLOYEE_ID
-----------
          2
          5

SQL>

还有一个选择：

 <snip>
 12  select employee_id
 13  from (select employee_id,
 14               case when skillset = 'c' then 1 else 0 end flag
 15        from employee_skillset
 16       )
 17  group by employee_id
 18  having sum(flag) = 0;

EMPLOYEE_ID
-----------
          2
          5

SQL>

或：

---

你考虑过为什么你会得到这些答案吗？基本上，您的查询首先选择skillset不是C的所有行，然后请求不同的ID。听起来好像你需要从另一端发起一些攻击。使用带有NOT EXISTS的sub query。为什么不使用下面的方法，因为如果员工还拥有其他技能，这将返回技能为“C”的员工。

 <snip>
 12  select employee_id
 13  from (select employee_id,
 14               listagg(skillset, ',') within group (order by null) lagg
 15        from employee_skillset
 16        group by employee_id
 17       )
 18  where instr(lagg, 'c') = 0;

EMPLOYEE_ID
-----------
          2
          5

SQL>