MYSQL IN

Question

美好的一天

我想查询在每个项目上工作的员工的姓名。 我的代码如下：

SELECT CONCAT (fname,' ', minit,'. ' , lname) AS NAME 
FROM employee a, works_on b, project c 
WHERE pno IN (Select pnumber  //WHAT COULD I SUBSTITUTE W/ IN 
       FROM projet) 
    AND a.ssn = b.essn 
    AND b.pno = c.pnumber 

在的问题是，它是价值观里面都像评价为“OR” ......是什么的，使我的子查询的评估值相当于like'AND'

谢谢您提前。

编辑： 作为reqeusted ..

Answer 1

SELECT * 
FROM employee e 
WHERE NOT EXISTS 
     (
     SELECT NULL 
     FROM employee ei 
     CROSS JOIN 
       project p 
     LEFT JOIN 
       works_on wo 
     ON  wo.pno = p.pnumber 
       AND wo.essn = ei.ssn 
     WHERE ei.ssn = e.ssn 
     ) 

Answer 2

select CONCAT (fname,' ', minit,'. ' , lname) AS NAME 
from employee 
left join works_on 
on works_on.essn=employee.ssn 
group by employee.ssn 
having count(works_on.essn) = (select count(*) from project); 

Answer 3

简化的例子：

create table emp_work_on 
(
emp_name varchar(50), 
work_on varchar(30) 
); 


create table works 
(
work_on varchar(30) 
); 

insert into works(work_on) values('apple'),('microsoft'),('google'),('facebook') 



insert into emp_work_on values 
('john','apple'),('john','microsoft'),('john','google'),('john','facebook'), 
('paul','microsoft'),('paul','google'), 
('george','apple'),('george','microsoft'),('george','google'),('george','facebook'), 
('ringo','apple'),('ringo','facebook'); 


select e.emp_name 
from works w 
left join emp_work_on e on e.work_on = w.work_on 
group by e.emp_name 
having count(e.work_on) = (select count(*) from works) 
order by e.emp_name 

输出：

emp_name 
---------- 
george 
john 
(2 rows) 

在你的表结构，你可以这样做：

SELECT * FROM employee 
WHERE ssn IN 
(
    SELECT w.essn 
    FROM project c 
    LEFT JOIN works_on w ON w.pno = c.pnumber 
    GROUP BY w.essn 
    HAVING COUNT(w.pno) = (SELECT COUNT(*) FROM project) 
) 

嗯..但我认为这可能是最简单的，给予有对员工的works_on没有重复PNO，即有一个在works_on那并不是没有PNO t存在于项目中，即保持参照完整性

SELECT * FROM employee 
WHERE ssn IN 
(
    SELECT essn 
    FROM works_on 
    GROUP BY essn 
    HAVING COUNT(pno) = (SELECT COUNT(*) FROM project) 
)