我正在尝试选择一名主管员工名单。
员工表：
+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+
| emp_id | fname | lname | start_date | end_date | superior_emp_id | dept_id | title | assigned_branch_id |
+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+
| 1 | Michael | Smith | 2005-06-22 | NULL | NULL | 3 | President | 1 |
| 2 | Susan | Barker | 2006-09-12 | NULL | 1 | 3 | Vice President | 1 |
| 3 | Robert | Tyler | 2005-02-09 | NULL | 1 | 3 | Treasurer | 1 |
| 4 | Susan | Hawthorne | 2006-04-24 | NULL | 3 | 1 | Operations Manager | 1 |
| 5 | John | Gooding | 2007-11-14 | NULL | 4 | 2 | Loan Manager | 1 |
| 6 | Helen | Fleming | 2008-03-17 | NULL | 4 | 1 | Head Teller | 1 |
| 7 | Chris | Tucker | 2008-09-15 | NULL | 6 | 1 | Teller | 1 |
| 8 | Sarah | Parker | 2006-12-02 | NULL | 6 | 1 | Teller | 1 |
| 9 | Jane | Grossman | 2006-05-03 | NULL | 6 | 1 | Teller | 1 |
| 10 | Paula | Roberts | 2006-07-27 | NULL | 4 | 1 | Head Teller | 2 |
| 11 | Thomas | Ziegler | 2004-10-23 | NULL | 10 | 1 | Teller | 2 |
| 12 | Samantha | Jameson | 2007-01-08 | NULL | 10 | 1 | Teller | 2 |
| 13 | John | Blake | 2004-05-11 | NULL | 4 | 1 | Head Teller | 3 |
| 14 | Cindy | Mason | 2006-08-09 | NULL | 13 | 1 | Teller | 3 |
| 15 | Frank | Portman | 2007-04-01 | NULL | 13 | 1 | Teller | 3 |
| 16 | Theresa | Markham | 2005-03-15 | NULL | 4 | 1 | Head Teller | 4 |
| 17 | Beth | Fowler | 2006-06-29 | NULL | 16 | 1 | Teller | 4 |
| 18 | Rick | Tulman | 2006-12-12 | NULL | 16 | 1 | Teller | 4 |
+--------+----------+-----------+------------+----------+-----------------+---------+--------------------+--------------------+
我可以通过此查询验证谁是主管：SELECT DISTINCT superior_emp_id FROM employee
我试图用这个选择一个只有主管的列表：
SELECT e.fname, e.lname, e.emp_id
FROM employee e
LEFT OUTER JOIN employee e2
ON e.superior_emp_id = e2.emp_id
ORDER BY e.fname
但我没有得到我想要的东西：
+----------+-----------+--------+
| fname | lname | emp_id |
+----------+-----------+--------+
| Beth | Fowler | 17 |
| Chris | Tucker | 7 |
| Cindy | Mason | 14 |
| Frank | Portman | 15 |
| Helen | Fleming | 6 |
| Jane | Grossman | 9 |
| John | Gooding | 5 |
| John | Blake | 13 |
| Michael | Smith | 1 |
| Paula | Roberts | 10 |
| Rick | Tulman | 18 |
| Robert | Tyler | 3 |
| Samantha | Jameson | 12 |
| Sarah | Parker | 8 |
| Susan | Hawthorne | 4 |
| Susan | Barker | 2 |
| Theresa | Markham | 16 |
| Thomas | Ziegler | 11 |
+----------+-----------+--------+
好吧，如果员工的ID出现在Employee.superior_emp_id列中，那么员工就被定义为主管。我们可以略微改写这个定义，如果Employee中至少有一行，他们的id在superior_emp_id列中，则说雇员是主管。这可以使用EXISTS查询来完成，并转换为以下SQL：
SELECT e.fname, e.lname, e.emp_id
FROM Employee AS e
WHERE EXISTS (
SELECT 1
FROM Employee e2
WHERE e2.superior_emp_id = e.emp_id
)
旁注：您实际上可以使用第一个语句编写查询。但是，有些优化器使用IN关键字非常困难 - 或者更糟糕的是如果涉及NULL - 并且通常我尽可能避免使用此关键字。为了完整性，虽然第一个声明将转换为：
SELECT e.fname, e.lname, e.emp_id
FROM Employee AS e
WHERE e.emp_id IN (SELECT superior_emp_id FROM employee)
另外：the self join suggested by @YogeshSharma也可以工作，前提是您选择上级（e1）的列而不是e中的员工。然而，这将为每个人产生多行，确切地说，每个人对应一个下行。这可以通过DISTINCT关键字或使用上面的查询来缓解。
试试这个查询
Select first_name, superior_emp_id, last_name from table where id= superior_emp_id
替换列名称将有助于您获得预期的数据