Question

问题是：列出与拥有empno 77的员工Joe Bloggs在同一电影院工作的员工的员工详细信息。

SELECT *
FROM Employee
WHERE Cinema_no IN (SELECT Cinema_no 
                    FROM Employee
                    WHERE Forename = 'Joe'
                      AND Surname = 'Bloggs')
  AND Empno <> 77;

没有错误消息，但是不会返回与Joe Bloggs在同一电影院工作的员工？

Answer 1

您的查询看起来还不错。我建议使用empno而不是surname / forname来识别员工。

SELECT *
FROM Employee
WHERE 
    Cinema_no IN (SELECT Cinema_no FROM Employee WHERE empno = 77)
    AND empno <> 77;

另一种解决方案是使用联接

SELECT emp.*
FROM Employee emp
JOIN ( SELECT Cinema_no FROM Employee WHERE empno = 77) cin on emp.cinema_no = cin.cinema_no
WHERE emp.empno <> 77;

Answer 2

Select * 
from employee 
where cinema_no in (select cimema_no 
                    from employees 
                    where emp_no = 77)
  and emp_no <> 77;

您的查询非常好，直到您拥有该姓氏/姓氏为止，如下所示

SELECT *
FROM Employee
WHERE Cinema_no IN (SELECT Cinema_no 
                    FROM Employee
                    WHERE UPPER(FORENAME) = UPPER('Joe')
                      AND UPPER(SURNAME) = UPPER('Bloggs'))
  AND empno <> 77

没有收到任何错误，但没有后续查询的数据？

2 个答案: