我有一张像这样有父子关系的桌子,
Employee_ID | Employee_Manager_ID | Employee_Name
--------------------------------------------------------
1 | 1 | AAAA
2 | 1 | BBBB
3 | 2 | CCCC
4 | 3 | DDDD
5 | 3 | EEEEE
是否可以使用单个查询将所有员工的人数归入特定员工(不仅是直接子女,孩子的所有孩子的数量)?
Eg if the input = 1
output should be 4
if input = 2 ,output should be 3
提前致谢
答案 0 :(得分:2)
假设您的表格为:
mysql> SELECT * FROM Employee;
+-----+------+-------------+------+
| SSN | Name | Designation | MSSN |
+-----+------+-------------+------+
| 1 | A | OWNER | 1 |
| 10 | G | WORKER | 5 |
| 11 | D | WORKER | 5 |
| 12 | E | WORKER | 5 |
| 2 | B | BOSS | 1 |
| 3 | F | BOSS | 1 |
| 4 | C | BOSS | 2 |
| 5 | H | BOSS | 2 |
| 6 | L | WORKER | 2 |
| 7 | I | BOSS | 2 |
| 8 | K | WORKER | 3 |
| 9 | J | WORKER | 7 |
+-----+------+-------------+------+
12 rows in set (0.00 sec)
查询:
SELECT SUPERVISOR.name AS SuperVisor,
GROUP_CONCAT(SUPERVISEE.name ORDER BY SUPERVISEE.name ) AS SuperVisee,
COUNT(*)
FROM Employee AS SUPERVISOR
INNER JOIN Employee SUPERVISEE ON SUPERVISOR.SSN = SUPERVISEE.MSSN
GROUP BY SuperVisor;
查询将产生如下结果:
+------------+------------+----------+
| SuperVisor | SuperVisee | COUNT(*) |
+------------+------------+----------+
| A | A,B,F | 3 |
| B | C,H,I,L | 4 |
| F | K | 1 |
| H | D,E,G | 3 |
| I | J | 1 |
+------------+------------+----------+
5 rows in set (0.00 sec)
[<强>答案强>]: 这对于一个级别(立即监督)来查找所有可能级别的所有监督,您必须在循环时使用(使用存储过程)。
虽然可以检索每个级别的员工然后采用他们的UNION,但我们通常不能指定查询,例如“在所有级别检索员工的受监管者”而不使用循环机制。 “
REFERENCE:在此幻灯片中读取滑动的数字23。 本书是“关系代数和关系微积分”一章中的“第四版数据库系统的基础”,主题是“递归闭包操作”。
为表创建添加查询可能对您有所帮助:
mysql> CREATE TABLE IF NOT EXISTS `Employee` (
-> `SSN` varchar(64) NOT NULL,
-> `Name` varchar(64) DEFAULT NULL,
-> `Designation` varchar(128) NOT NULL,
-> `MSSN` varchar(64) NOT NULL,
-> PRIMARY KEY (`SSN`),
-> CONSTRAINT `FK_Manager_Employee` FOREIGN KEY (`MSSN`) REFERENCES Employee(SSN)
-> ) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Query OK, 0 rows affected (0.17 sec)
您可以查看表格:
mysql> DESCRIBE Employee;
+-------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------------+--------------+------+-----+---------+-------+
| SSN | varchar(64) | NO | PRI | NULL | |
| Name | varchar(64) | YES | | NULL | |
| Designation | varchar(128) | NO | | NULL | |
| MSSN | varchar(64) | NO | MUL | NULL | |
+-------------+--------------+------+-----+---------+-------+
4 rows in set (0.00 sec)
答案 1 :(得分:0)
你可以试试这个:
SELECT
table_name.Employee_ID,
table_name.Employee_Name,
COUNT(*) AS children
FROM
table_name AS t_one
INNER JOIN table_name AS t_two ON
t_two.Employee_Manager_ID=t_one.Employee_ID
GROUP BY
t_one.Employee_ID