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 EI 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 share|improve this question edited Dec 11 '12 at 9:28 asked Dec 6 '12 at 15:36
Grijesh Chauhan
2862413
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
| show 11 more comments
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.
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
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>
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
add a comment |
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.
share|improve this answer 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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