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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!


mysql sql stored-procedures    
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edited Dec 11 '12 at 9:28    
 
 

        asked Dec 6 '12 at 15:36    

    
Grijesh Chauhan        2862413        



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

This question came from our site for professional and enthusiast programmers.


 
1                                                                                  
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                                                                            

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2 Answers                                 2

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         up vote2down 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 latin1DETERMINISTICBEGIN  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>

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edited Dec 11 '12 at 17:11    
 
 

        answered Dec 10 '12 at 20:09    

    
RolandoMySQLDBA        102k13131263        


 
                                                                                                                   
Excellent ...Thanks A Lots!                     – Grijesh Chauhan                 Dec 12 '12 at 9:11                                                                            
                                                                                                                   
understood the Idea!                     – Grijesh Chauhan                 Dec 12 '12 at 9:15                                                                            

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         up vote2down 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.


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        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                                                                            

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protected by RolandoMySQLDBA Dec 11 '12 at 18:38

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