MySQL: Tree-Hierarchical query

 

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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!

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

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Sample test fiddle                     – mellamokb                 Dec 6 '12 at 15:46                                                                            
                                                                                                                    
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  Thanks mellamokb ! :)                     – Grijesh Chauhan                 Dec 6 '12 at 15:52                                                                            
1                                                                                  
@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                                                                            
1                                                                                  
@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                                                                            

2 Answers                                 2

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         up vote 2down voteaccepted

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 <>

转载于:https://www.cnblogs.com/kungfupanda/p/5645607.html

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