显示引用的名称而不是同一个表中的ID

时间:2012-07-16 15:45:02

标签: sql

我有一个名为employee的表,其中包含一个ID为Primary Key的列和一个名为supervisorID的列,该列是对引用某个人的同一个表的引用。我想将supervisorID显示为其引用而不是ID的人名。

我想从表employee中选择*,但是将SupervisorID作为同一个表中的人名引用。

3 个答案:

答案 0 :(得分:7)

SELECT e.ID, e.name AS Employee, s.name AS Supervisor 
FROM employee e 
  INNER JOIN employee s 
  ON s.ID = e.supervisorID 
ORDER BY e.ID;

以下是有关如何测试的更多颜色:

mysql> CREATE TABLE employee (ID INT NOT NULL AUTO_INCREMENT, supervisorID INT NOT NULL DEFAULT '1', name VARCHAR(48) NOT NULL, PRIMARY KEY (ID));
Query OK, 0 rows affected (0.01 sec)


mysql> INSERT INTO employee VALUES (1, 1, "The Boss"), (2,1, "Some Manager"), (3,2, "Some Worker"), (4,2, "Another Worker");
Query OK, 4 rows affected (0.00 sec)
Records: 4  Duplicates: 0  Warnings: 0


mysql> SELECT e.ID, e.name AS Employee, s.name AS Supervisor
FROM employee e INNER JOIN employee s
ON s.ID = e.supervisorID ORDER BY e.ID;
+----+----------------+--------------+
| ID | Employee       | Supervisor   |
+----+----------------+--------------+
|  1 | The Boss       | The Boss     |
|  2 | Some Manager   | The Boss     |
|  3 | Some Worker    | Some Manager |
|  4 | Another Worker | Some Manager |
+----+----------------+--------------+
4 rows in set (0.01 sec)

mysql> 

答案 1 :(得分:0)

类似的东西:

SELECT e.name as 'employee name', supervisors.name as 'supervisor name'
FROM employee e
INNER JOIN employee supervisors ON e.ID = supervisors.supervisorID

答案 2 :(得分:0)

根据ID

进行自我加入
select e.*, s.name
from 
  employee e
  inner join employee s
    on e.supervisorid = s.id