SUB-TREE WITHIN A TREE in MySQL
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In my MYSQL
Database COMPANY, I have a Table: Employee with recursive association, an employee can be boss of other employee. A self relationship of kind (SuperVisor (1)- SuperVisee (∞) ).
Query to Create Table:
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;
I have inserted a set of tuples (Query):
INSERT INTO Employee VALUES
("1", "A", "OWNER", "1"),
("2", "B", "BOSS", "1"), # Employees under OWNER
("3", "F", "BOSS", "1"),
("4", "C", "BOSS", "2"), # Employees under B
("5", "H", "BOSS", "2"),
("6", "L", "WORKER", "2"),
("7", "I", "BOSS", "2"),
# Remaining Leaf nodes
("8", "K", "WORKER", "3"), # Employee under F
("9", "J", "WORKER", "7"), # Employee under I
("10","G", "WORKER", "5"), # Employee under H
("11","D", "WORKER", "4"), # Employee under C
("12","E", "WORKER", "4")
The inserted rows has following **Tree-Hierarchical-Relationship**:
A <---ROOT-OWNER
/|\
/ A \
B F
//| \ \
// | \ K
/ | | \
I L H C
/ | / \
J G D E
I written a query to find relationship:
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;
And output is:
+------------+------------+----------+
| SuperVisor | SuperVisee | COUNT(*) |
+------------+------------+----------+
| A | A,B,F | 3 |
| B | C,H,I,L | 4 |
| C | D,E | 2 |
| F | K | 1 |
| H | G | 1 |
| I | J | 1 |
+------------+------------+----------+
6 rows in set (0.00 sec)
[**QUESTION**]
Instead of complete Hierarchical Tree, I need a SUB-TREE from a point (selective) e.g.:
If input argument is B then output should be as below...
+------------+------------+----------+
| SuperVisor | SuperVisee | COUNT(*) |
+------------+------------+----------+
| B | C,H,I,L | 4 |
| C | D,E | 2 |
| H | G | 1 |
| I | J | 1 |
+------------+------------+----------+
Please help me on this. If not query, a stored-procedure can be helpful.
I tried, but all efforts were useless!
Asked by Grijesh Chauhan
(581 rep)
Dec 6, 2012, 03:36 PM
Last activity: Jun 27, 2025, 03:24 PM
Last activity: Jun 27, 2025, 03:24 PM