How to Update same table on deletion in MYSQL?

In my Database I have a table: Employee with recursive association, an employee can be boss of other employee.

Following is query I used 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) DEFAULT NULL, 
  PRIMARY KEY (`SSN`),
  CONSTRAINT `FK_Manager_Employee`  FOREIGN KEY (`MSSN`) REFERENCES Employee(SSN)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

And I could successfully create my table:

mysql> DESCRIBE Employee;
+-------------+--------------+------+-----+---------+-------+
| Field       | Type         | Null | Key | Default | Extra |
+-------------+--------------+------+-----+---------+-------+
| SSN         | varchar(64)  | NO   | PRI | NULL    |       |
| name        | varchar(64)  | YES  |     | NULL    |       |
| designation | varchar(128) | NO   |     | NULL    |       |
| MSSN        | varchar(64)  | YES  | MUL | NULL    |       |
+-------------+--------------+------+-----+---------+-------+
4 rows in set (0.00 sec)

Then I inserted some tuple in Employee table using following query.

mysql> INSERT INTO Employee VALUES 
    -> ("1", "A", "OWNER",  NULL), 
    -> ("2", "B", "BOSS",   "1"),
    -> ("3", "C", "WORKER", "2"),
    -> ("4", "D", "BOSS",   "2"),
    -> ("5", "E", "WORKER", "4"),
    -> ("6", "F", "WORKER", "1"),
    -> ("7", "G", "WORKER", "4")
    -> ;
Query OK, 7 rows affected (0.02 sec)
Records: 7  Duplicates: 0  Warnings: 0

Now I have following hierarchical relation among the rows in table:

A
       / \
      B   F
     / \
    C   D
       / \
      G   E

Following is Select statement for table:

```
mysql> SELECT * FROM Employee;
+-----+------+-------------+------+
| SSN | name | designation | MSSN |
+-----+------+-------------+------+
| 1 | A | OWNER | NULL |
| 2 | B | BOSS | 1 |
| 3 | C | WORKER | 2 |
| 4 | D | B

The easiest way to delete D is to do the following:

• Have B inherit G and E, putting them on the same level as D.

• Delete D

First, I'll load your sample data

mysql> DROP DATABASE IF EXISTS grijesh;
Query OK, 1 row affected (0.06 sec)

mysql> CREATE DATABASE grijesh;
Query OK, 1 row affected (0.00 sec)

mysql> USE grijesh
Database changed
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) DEFAULT NULL,
    ->   PRIMARY KEY (`SSN`),
    ->   CONSTRAINT `FK_Manager_Employee`  FOREIGN KEY (`MSSN`) REFERENCES Employee(SSN)
    -> ) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Query OK, 0 rows affected (0.08 sec)

mysql> INSERT INTO Employee VALUES
    -> ("1", "A", "OWNER",  NULL),
    -> ("2", "B", "BOSS",   "1"),
    -> ("3", "C", "WORKER", "2"),
    -> ("4", "D", "BOSS",   "2"),
    -> ("5", "E", "WORKER", "4"),
    -> ("6", "F", "WORKER", "1"),
    -> ("7", "G", "WORKER", "4");
Query OK, 7 rows affected (0.05 sec)
Records: 7  Duplicates: 0  Warnings: 0

mysql> SELECT * FROM Employee;
+-----+------+-------------+------+
| SSN | name | designation | MSSN |
+-----+------+-------------+------+
| 1   | A    | OWNER       | NULL |
| 2   | B    | BOSS        | 1    |
| 3   | C    | WORKER      | 2    |
| 4   | D    | BOSS        | 2    |
| 5   | E    | WORKER      | 4    |
| 6   | F    | WORKER      | 1    |
| 7   | G    | WORKER      | 4    |
+-----+------+-------------+------+
7 rows in set (0.00 sec)

mysql>

View all current relationships

mysql> SELECT
    ->     A.name,A.designation,
    ->     B.name,B.designation
    -> FROM
    ->     `Employee` A INNER JOIN `Employee` B ON A.SSN= B.MSSN
    -> ORDER BY A.SSN,B.SSN
    -> ;
+------+-------------+------+-------------+
| name | designation | name | designation |
+------+-------------+------+-------------+
| A    | OWNER       | B    | BOSS        |
| A    | OWNER       | F    | WORKER      |
| B    | BOSS        | C    | WORKER      |
| B    | BOSS        | D    | BOSS        |
| D    | BOSS        | E    | WORKER      |
| D    | BOSS        | G    | WORKER      |
+------+-------------+------+-------------+
6 rows in set (0.00 sec)

mysql>

With this UPDATE JOIN query

UPDATE `Employee` A INNER JOIN `Employee` B ON A.SSN= B.MSSN
SET B.MSSN = A.MSSN WHERE A.SSN = '4';

let's elevate G and E to the same level as D and show relationships

mysql> UPDATE `Employee` A INNER JOIN `Employee` B ON A.SSN= B.MSSN
    -> SET B.MSSN = A.MSSN WHERE A.SSN = '4';
Query OK, 2 rows affected (0.06 sec)
Rows matched: 2  Changed: 2  Warnings: 0

mysql> SELECT
    ->     A.name,A.designation,
    ->     B.name,B.designation
    -> FROM
    ->     `Employee` A INNER JOIN `Employee` B ON A.SSN= B.MSSN
    -> ORDER BY A.SSN,B.SSN
    -> ;
+------+-------------+------+-------------+
| name | designation | name | designation |
+------+-------------+------+-------------+
| A    | OWNER       | B    | BOSS        |
| A    | OWNER       | F    | WORKER      |
| B    | BOSS        | C    | WORKER      |
| B    | BOSS        | D    | BOSS        |
| B    | BOSS        | E    | WORKER      |
| B    | BOSS        | G    | WORKER      |
+------+-------------+------+-------------+
6 rows in set (0.00 sec)

mysql>

At this point, D is nobody's boss.

Now, delete it and show relationships:

mysql> DELETE FROM `Employee` WHERE SSN = '4';
Query OK, 1 row affected (0.05 sec)

mysql> SELECT
    ->     A.name,A.designation,
    ->     B.name,B.designation
    -> FROM
    ->     `Employee` A INNER JOIN `Employee` B ON A.SSN= B.MSSN
    -> ORDER BY A.SSN,B.SSN
    -> ;
+------+-------------+------+-------------+
| name | designation | name | designation |
+------+-------------+------+-------------+
| A    | OWNER       | B    | BOSS        |
| A    | OWNER       | F    | WORKER      |
| B    | BOSS        | C    | WORKER      |
| B    | BOSS        | E    | WORKER      |
| B    | BOSS        | G    | WORKER      |
+------+-------------+------+-------------+
5 rows in set (0.00 sec)

mysql>

Give it a Try !!!

Please notice that you do not need a Trigger

StackExchange Database Administrators Q#29047, answer score: 2