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 (∞) ).

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!

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

score 2down voteaccepted · Accepted Answer

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

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 <>

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

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