SQL问题-如何让员工拥有与特定员工相同的技能

Question

我有一个名为Employee的表，其中包含列：

• EMPID INT(主键(
• EMPNAME VARCHAR2(100(

样本数据如下：

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

我有另一个带有列的表Employee_Skills：

• EMPID INT
• 技能变量2(100(

其样本可表示如下：

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

在第二个表中，这两列一起构成主键。并且EMPID具有指向Employee(EMPID)的引用约束。

我想让所有的员工都具备玛丽的所有技能。

例如，

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

此查询返回以下行：

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

现在，我想选择所有具备上述技能的员工(可以拥有更多技能，但至少应该拥有Mary拥有的技能(。

我做了几次尝试，但没有什么能满足我的需要。

有一次尝试没有成功：

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');

Answer 1

您可以通过联接和聚合来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'
)

Answer 2

有很多方法可以解决这个问题，其中最直观的方法可能是：

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);

附言：我把玛丽排除在返回的名单之外。

我重新创建了您的数据如下：

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');          

Answer 3

这已经奏效了。为每个人分享。基本上，你需要玛丽的技能。对当前员工的技能做一分钟评估。如果您得到一个空白，则在结果中显示该员工。

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';