SQL 问题 - 如何让员工拥有与特定员工相同的技能

SQL question - How to get employees with the same skills as a specific employee

我有一个名为 Employee 的 table,其列为:

样本数据如下:

empid|empname|
-----|-------|
    1|Mary   |
    2|Bob    |

我有另一个 table Employee_Skills 列:

其中的一个样本可以表示如下:

empid|skill |
-----|------|
    1|SKILL |
    1|Java  |
    1|C++   |
    1|Python|
    2|C++   |
    2|Python|

在第二个table中,两列共同构成了主键。 EMPID 有一个指向 Employee(EMPID).

的引用约束

我希望所有员工都具备 Mary 所拥有的所有技能。

例如,

select t2.skill 
from Employee t1 
join Employee_Skills t2 on (t1.EMPID = t2.EMPID) 
where t1.EMPNAME = 'Mary'

此查询返回了这些行:

SKILL
------------
Java
C++
Python
... (a few more)

现在我想 select 所有具备上述技能的员工(可以有更多的技能,但至少应该有 Mary 拥有的技能)。

我已经做了几次尝试,但没有任何东西能满足我的需求。

这是一种失败的尝试:

select t1.EMPID, t1.EMPNAME
from Employees t1 
join Employee_Skills t2 on (t1.EMPID = t2.EMPID)
where t2.SKILL = ALL (select t4.skill 
                      from Employee t3  
                      join Employee_Skills t4 on (t3.EMPID = t4.EMPID) 
                      where t3.EMPNAME = 'Mary');

你可以用连接和聚合解决这个关系划分问题:

select e2.empid
from employee e1
inner join employee_skills es1 on es1.empid = e1.empid
inner join employee_skills es2 on es2.skill = es1.skill and es2.empid <> es1.empid 
inner join employee e2 on e2.empid = es2.empid
where e1.empname = 'Mary'
group by e2.empid
having count(*) = (
    select count(*)
    from employee e3
    inner join employee_skills es3 on es3.empid = e3.empid
    where e3.empname = 'Mary'
)

有很多方法可以解决这个问题,其中最直观的一种可能是以下方法:

select emp.empid, emp.empname
from Employee_Skills esk
inner join employee emp
 on emp.empid = esk.empid
where emp.empname <> 'Mary' 
 and exists (select null   
                from employee_skills esk_
                 inner join employee emp_
                  on emp_.empid = esk_.empid
                where emp_.empname = 'Mary'
                 and esk_.skill = esk.skill);

PS: 我从返回的列表中排除了 Mary。

我重新创建了你的数据如下:

create table employee(EMPID INT , EMPNAME VARCHAR(100));
                                                   
create table Employee_Skills(EMPID INT, SKILL VARCHAR(100));
                                                       
insert into employee values (1, 'Mary');
insert into employee values (2, 'Bob');
                                                       
insert into employee_skills values (2, 'SKILL');
insert into employee_skills values (2, 'Java');
insert into employee_skills values (1, 'C++');
insert into employee_skills values (1, 'Python');                                 
insert into employee_skills values (2, 'C++');
insert into employee_skills values (2, 'Python');          
                

这成功了。分享给大家。基本上,你会掌握玛丽的技能。减去当前员工的技能。如果你得到一个空白,你就会在结果中显示这个员工。

Select t4.empid, t4.empname from employee t4
Where not exists 
(Select t1.skill from employee_skills t1 join employee t2 on (t1.empid = t2.empid) where t2.empname = 'Mary'
Minus
Select t3.skill from employee_skills t3 where t3.empid = t4.empid)
and t4.empname <> 'Mary';