MySQL: Tree-Hierarchical query

Question

SUB-TREE WITHIN A TREE in MySQL

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 (∞) ). node

Query to Create Table: 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;

I have inserted a set of tuples (Query): web

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: sql

 A <---ROOT-OWNER /|\ / A \ B F //| \ \ // | \ K / | | \ I L H C / | / \ J G D E

I written a query to find relationship: app

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: less

+------------+------------+----------+ | 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...dom

+------------+------------+----------+ | 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!ide

I simply provided a test framework for the community to use in exploring this question more easily. – mellamokb Dec 6 '12 at 15:50 — mellamokb, Dec 6 '12 at 15:50
@mellamokb Thanks mellamokb ! :) – Grijesh Chauhan Dec 6 '12 at 15:52 — Grijesh Chauhan, Dec 6 '12 at 15:52
@GrijeshChauhan let me ask you this: Which is better to make your own visible waves? To throw pebbles into the ocean, or to throw rocks into a small pond? Going straight to the experts is almost certainly going to give you the best answer, and this sort of question is so important (advanced database topics) that we have given it its own site on the network. But I won't stop you from asking it where you like, that's your prerogative. My prerogative is to vote to move it to another site if I think that's where it belongs. :D We both use the network as we see fit in this case :D – jcolebrand♦ Dec 6 '12 at 16:33 — jcolebrand♦, Dec 6 '12 at 16:33
@jcolebrand: Actually it was my fault only. I use to post question on multiple sides to get a better, quick and many response. It my experience I always got better answer from expert sides. And I think it was better decision to move question to Database Administrators. In all the cases, I am very thankful to stackoverflow and peoples who are active here. I really got solution for many problem that was very tough to find myself or any other web. – Grijesh Chauhan Dec 6 '12 at 16:43 — Grijesh Chauhan, Dec 6 '12 at 16:43

RolandoMySQLDBA · Accepted Answer · 2012-12-11 17:11:15Z

I already addressed something of this nature using Stored Procedures : Find highest level of a hierarchical field: with vs without CTEs (Oct 24, 2011)ui

If you look in my post, you could use the GetAncestry and GetFamilyTree functions as a model for traversing the tree from any given point.

UPDATE 2012-12-11 12:11 EDT

I looked back at my code from my post. I wrote up the Stored Function for you:

DELIMITER $$ DROP FUNCTION IF EXISTS `cte_test`.`GetFamilyTree` $$ CREATE FUNCTION `cte_test`.`GetFamilyTree`(GivenName varchar(64)) RETURNS varchar(1024) CHARSET latin1 DETERMINISTIC BEGIN DECLARE rv,q,queue,queue_children,queue_names VARCHAR(1024); DECLARE queue_length,pos INT; DECLARE GivenSSN,front_ssn VARCHAR(64); SET rv = ''; SELECT SSN INTO GivenSSN FROM Employee WHERE name = GivenName AND Designation <> 'OWNER'; IF ISNULL(GivenSSN) THEN RETURN ev; END IF; SET queue = GivenSSN; SET queue_length = 1; WHILE queue_length > 0 DO IF queue_length = 1 THEN SET front_ssn = queue; SET queue = ''; ELSE SET pos = LOCATE(',',queue); SET front_ssn = LEFT(queue,pos - 1); SET q = SUBSTR(queue,pos + 1); SET queue = q; END IF; SET queue_length = queue_length - 1; SELECT IFNULL(qc,'') INTO queue_children FROM ( SELECT GROUP_CONCAT(SSN) qc FROM Employee WHERE MSSN = front_ssn AND Designation <> 'OWNER' ) A; SELECT IFNULL(qc,'') INTO queue_names FROM ( SELECT GROUP_CONCAT(name) qc FROM Employee WHERE MSSN = front_ssn AND Designation <> 'OWNER' ) A; IF LENGTH(queue_children) = 0 THEN IF LENGTH(queue) = 0 THEN SET queue_length = 0; END IF; ELSE IF LENGTH(rv) = 0 THEN SET rv = queue_names; ELSE SET rv = CONCAT(rv,',',queue_names); END IF; IF LENGTH(queue) = 0 THEN SET queue = queue_children; ELSE SET queue = CONCAT(queue,',',queue_children); END IF; SET queue_length = LENGTH(queue) - LENGTH(REPLACE(queue,',','')) + 1; END IF; END WHILE; RETURN rv; END $$

It actually works. Here is a sample:

mysql> SELECT name,GetFamilyTree(name) FamilyTree -> FROM Employee WHERE Designation <> 'OWNER'; +------+-----------------------+ | name | FamilyTree | +------+-----------------------+ | A | B,F,C,H,L,I,K,D,E,G,J | | G | | | D | | | E | | | B | C,H,L,I,D,E,G,J | | F | K | | C | D,E | | H | G | | L | | | I | J | | K | | | J | | +------+-----------------------+ 12 rows in set (0.36 sec) mysql>

There is only one catch. I added one extra row for the owner

The owner has SSN 0
The owner is his own boss with MSSN 0

Here is the data

mysql> select * from Employee; +-----+------+-------------+------+ | SSN | Name | Designation | MSSN | +-----+------+-------------+------+ | 0 | A | OWNER | 0 | | 1 | A | BOSS | 0 | | 10 | G | WORKER | 5 | | 11 | D | WORKER | 4 | | 12 | E | WORKER | 4 | | 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 | +-----+------+-------------+------+ 13 rows in set (0.00 sec) mysql>

Excellent ...Thanks A Lots! – Grijesh Chauhan Dec 12 '12 at 9:11 — Grijesh Chauhan, Dec 12 '12 at 9:11

Shiplu · Answer 2 · 2012-12-06 15:46:18Z

up vote 2down vote

What you are using is called Adjacency List Model. It has a lot of limitations. You'll be problem when you want to delete/insert a node at a specific place. Its better you use Nested Set Model.

There is a detailed explanation. Unfortunately the article on mysql.com is does not exist any more.

answered Dec 6 '12 at 15:46

Shiplu

1213

5

"it has a lot of limitations" - but only when using MySQL. Nearly all DBMS support recursive queries (MySQL is one of the very few that doesn't) and that makes the model really easy to deal with. – a_horse_with_no_name Dec 7 '12 at 7:05

@a_horse_with_no_name Never used anything other than MySQL. So I never knew it. Thanks for the information. – Shiplu Dec 7 '12 at 11:15

add a comment |

MySQL: Tree-Hierarchical query

SUB-TREE WITHIN A TREE in MySQL

migrated from stackoverflow.com Dec 8 '12 at 9:42

2 Answers 2

UPDATE 2012-12-11 12:11 EDT

protected by RolandoMySQLDBA Dec 11 '12 at 18:38