Insert rows into the MySQL 'employees' table with 'department_id' and 'manager_id' values ensuring a unique combination exists in the 'departments' table

MySQL insert into Statement: Exercise-13 with Solution

13. Write a MySQL query to insert rows into the table employees in which a set of columns department_id and manager_id contains a unique value and that combined values must have exists into the table departments.

Sample table departments.


CREATE TABLE IF NOT EXISTS departments ( 
DEPARTMENT_ID integer NOT NULL UNIQUE, 
DEPARTMENT_NAME varchar(30) NOT NULL, 
MANAGER_ID integer DEFAULT NULL, 
LOCATION_ID integer DEFAULT NULL, 
PRIMARY KEY (DEPARTMENT_ID,MANAGER_ID) 
)ENGINE=InnoDB;


INSERT INTO departments VALUES(60,'SALES',201,89);
INSERT INTO departments VALUES(61,'ACCOUNTS',201,89);
INSERT INTO departments VALUES(80,'FINANCE',211,90);

mysql> SELECT * FROM departments;
+---------------+-----------------+------------+-------------+
| DEPARTMENT_ID | DEPARTMENT_NAME | MANAGER_ID | LOCATION_ID |
+---------------+-----------------+------------+-------------+
|            60 | SALES           |        201 |          89 |
|            61 | ACCOUNTS        |        201 |          89 |
|            80 | FINANCE         |        211 |          90 |
+---------------+-----------------+------------+-------------+
3 rows in set (0.00 sec)

Sample table employees.

CREATE TABLE IF NOT EXISTS employees ( 
EMPLOYEE_ID integer NOT NULL PRIMARY KEY, 
FIRST_NAME varchar(20) DEFAULT NULL, 
LAST_NAME varchar(25) NOT NULL, 
JOB_ID varchar(10) NOT NULL, 
SALARY decimal(8,2) DEFAULT NULL, 
MANAGER_ID integer DEFAULT NULL, 
DEPARTMENT_ID integer DEFAULT NULL, 
FOREIGN KEY(DEPARTMENT_ID,MANAGER_ID) 
REFERENCES  departments(DEPARTMENT_ID,MANAGER_ID)
)ENGINE=InnoDB;

Now insert the rows in the employees.

Sample Solution:

-- Inserting a new record into the 'employees' table
INSERT INTO employees VALUES(510, 'Alex', 'Hanes', 'CLERK', 18000, '201', 60);
-- Inserting another record into the 'employees' table
INSERT INTO employees VALUES(511, 'Kim', 'Leon', 'CLERK', 18000, '211', 80);

Let execute the above code in MySQL 5.6 command prompt.

Here is the structure of the table:

mysql> SELECT * FROM employees;
+-------------+------------+-----------+--------+----------+------------+---------------+
| EMPLOYEE_ID | FIRST_NAME | LAST_NAME | JOB_ID | SALARY   | MANAGER_ID | DEPARTMENT_ID |
+-------------+------------+-----------+--------+----------+------------+---------------+
|         510 | Alex       | Hanes     | CLERK  | 18000.00 |        201 |            60 |
|         511 | Kim        | Leon      | CLERK  | 18000.00 |        211 |            80 |
+-------------+------------+-----------+--------+----------+------------+---------------+
2 rows in set (0.00 sec)

The value against department_id and manager_id combination (60,201) and (80,211) are unique in the departmentis(parent) table so, there is no problem arise to insert the rows in the child table employees.

Now insert another row in the employees table.

INSERT INTO employees VALUES(512,'Kim','Leon','CLERK',18000,80,211);

Let execute the above code in MySQL command prompt.

mysql> INSERT INTO employees VALUES(512,'Kim','Leon','CLERK',18000,80,211);
ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (`hrr`.`employees`, CONSTRAINT `employees_ibfk_1` FOREIGN KEY (`D
EPARTMENT_ID`, `MANAGER_ID`) REFERENCES `departments` (`DEPARTMENT_ID`, `MANAGER_ID`))

Here in the above, the value against department_id and manager_id combination (211,80) does not matching with the same combination in departments(parent table) table and that is why the child table employees can not contain the combination of values including department_id and manager_id as specified. Here the primary key - foreign key relationship is being violated and shows the above message.

Explanation:

Here's a breakdown of the above MySQL query:

Insert a new record into the 'employees' table with the following values:

• EMPLOYEE_ID: 510
• FIRST_NAME: 'Alex'
• LAST_NAME: 'Hanes'
• JOB_ID: 'CLERK'
• SALARY: 18000
• MANAGER_ID: '201'
• DEPARTMENT_ID: 6

Insert another new record into the 'employees' table with similar columns and different values:

• EMPLOYEE_ID: 511
• FIRST_NAME: 'Kim'
• LAST_NAME: 'Leon'
• JOB_ID: 'CLERK'
• SALARY: 18000
• MANAGER_ID: '211'
• DEPARTMENT_ID: 80

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