In Oracle SQL / PLSQL a FOREIGN KEY is column / field that appears in one table and must appear in another table.
Important points about FOREIGN KEY ON DELETE SET NULL in Oracle SQL / PLSQL:
- A FOREIGN KEY creates a parent child pattern between two tables.
- The referenced table is called the PARENT table and the table having the FOREIGN KEY is called as the CHILD table.
- The FOREIGN KEY in child table generally references PRIMARY KEY in parent table.
- A foreign key with cascade delete can be defined in CREATE TABLE or ALTER TABLE statement.
A FOREIGN KEY with ON DELETE SET NULL means that whenever a record from the parent table is deleted all the corresponding record(s) in the child table will have the FOREIGN KEY fields set to NULL. The records from the child table are not deleted unlike ON DELETE CASCADE.
Syntax to create a FOREIGN KEY with ON DELETE SET NULL in CREATE statement is:
CREATE TABLE table_name
(column_name1 datatype null/NOT NULL
,column_name2 datatype null/NOT NULL
,column_name3 datatype null/NOT NULL
.
.
,column_nameN datatype null/NOT NULL
,CONSTRAINT constraint_name
FOREIGN KEY (column_name1, column_name2, . . . column_nameN )
REFERENCES parent_table_name(column_name1, column_name2, . . . column_nameN)
ON DELETE SET NULL
);
Syntax to create a FOREIGN KEY with ON DELETE SET NULL in ALTER statement is:
ALTER TABLE TABLE_NAME
ADD CONSTRAINT constraint_name
FOREIGN KEY (column_name1,column_name2, . . column_nameN)
REFERENCES parent_table_name (column_name1,column_name2, . . column_nameN)
ON DELETE SET NULL;
Let’s take an example for understanding:
Scenario 1:
Step 1: Suppose we want to create a table named ‘employee’ in the database as shown below, with ‘employee_id’ as the primary key for the ‘employee’ table
Employee_Id | Employee_Name | Salary | Department |
101 | Emp A | 10000 | Sales |
102 | Emp B | 20000 | IT |
103 | Emp C | 28000 | IT |
104 | Emp D | 30000 | Support |
105 | Emp E | 32000 | Sales |
And we want to create a table named ‘comm’ with ‘emp_id’ as FOREIGN KEY as shown below:
Emp_Id | Commission_Percent |
102 | 20 |
103 | 20 |
104 | |
105 | 10 |
We can achieve the same as:
CREATE TABLE employee (employee_id NUMBER(10) NOT NULL ,employee_name VARCHAR2(500) NOT NULL ,salary NUMBER(20) NOT NULL ,department VARCHAR2(300) NOT NULL ,CONSTRAINT employee_pk PRIMARY KEY (employee_id) );
CREATE TABLE comm (emp_id NUMBER(10) ,commission_percent NUMBER(20) ,CONSTRAINT fk_employee FOREIGN KEY (emp_id) REFERENCES employee(employee_id) ON DELETE SET NULL );
Here with the help of the above SQL CREATE statement we have created a table named ‘employee’ that has 4 columns namely ‘employee_id’, ‘employee_name’, ‘department’ and ‘salary’ and we are having the ‘employee_id’ column as the primary key for the ‘employee’ table.
Also, we have created a new table named ‘comm’ which has 2 columns namely ‘emp_id’ and ‘commission_percent’ and we are having ‘emp_id’ as foreign key referencing ‘employee_id’ of ‘employee’ table.
Step 2: Inserting the data in the ‘employee’ table.
INSERT INTO employee VALUES (101,'Emp A',10000,'Sales'); INSERT INTO employee VALUES (102,'Emp B',20000,'IT'); INSERT INTO employee VALUES (103,'Emp C',28000,'IT'); INSERT INTO employee VALUES (104,'Emp D',30000,'Support'); INSERT INTO employee VALUES (105,'Emp E',32000,'Sales');
Step 3: Inserting the data in the ‘comm’ table.
INSERT INTO comm VALUES (102,20); INSERT INTO comm VALUES (103,20); INSERT INTO comm VALUES (104,NULL); INSERT INTO comm VALUES (105,10);
The above SQL INSERT statements will insert 5 rows in the ‘employee’ table and 4 rows in ‘comm’ table.
Now, if we query the employee table as:
SELECT * FROM employee;
We will get the following result:
Employee_Id | Employee_Name | Salary | Department |
101 | Emp A | 10000 | Sales |
102 | Emp B | 20000 | IT |
103 | Emp C | 28000 | IT |
104 | Emp D | 30000 | Support |
105 | Emp E | 32000 | Sales |
Now, if we query the ‘comm’ table as:
SELECT * FROM comm;
We will get the following result:
Emp_Id | Commission_Percent |
102 | 20 |
103 | 20 |
104 | |
105 | 10 |
Here we have successfully created ‘employee’ and ‘comm’ tables with ‘employee_id’ as PRIMARY KEY for ‘employee’ table and ‘emp_id’ as foreign key for ‘comm’ and also inserted data in ‘employee’ and ‘comm’ tables.
If we try to insert one more record in ‘comm’ table of ‘emp_id = 106’ as:
INSERT INTO comm VALUES (106,30);
We get an error that integrity constraint is getting violated and the data cannot be entered, this is because ‘emp_id = 106’ is not found in ‘employee’ table and hence the record cannot be entered in ‘comm’ table.
Lets delete a record of ‘employee_id =105’ from ‘employee’ table
DELETE FROM employee WHERE employee_id = 105;
The above statement will delete a record of ‘employee_id = 105’ from ‘employee’ table and when we query the ‘employee’ table as
SELECT * FROM employee;
We will get the below result
Employee_Id | Employee_Name | Salary | Department |
101 | Emp A | 10000 | Sales |
102 | Emp B | 20000 | IT |
103 | Emp C | 28000 | IT |
104 | Emp D | 30000 | Support |
Also if we query the ‘comm’ table as
SELECT * FROM comm;
We get the below result:
Emp_Id | Commission_Percent |
102 | 20 |
103 | 20 |
104 | |
10 |
We can see that the record for ‘emp_id = 105’ has been set to NULL in the ‘comm’ table because we have written ON DELETE SET NULL while creating FOREIGN KEY in ‘comm’ table.
Scenario 2:
Using ALTER TABLE statement to create a FOREIGN KEY with ON DELETE CASCADE
Syntax for creating a FOREIGN KEY using ALTER TABLE statement is:
ALTER TABLE TABLE_NAME
ADD CONSTRAINT constraint_name
FOREIGN KEY (column_name1,column_name2, . . column_nameN)
REFERENCES parent_table_name (column_name1,column_name2, . . column_nameN)
ON DELETE SET NULL;
Example:
ALTER TABLE comm ADD CONSTRAINT comm_fk FOREIGN KEY (emp_id) REFERENCES employee (employee_id) ON DELETE SET NULL;