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PostgreSQL Create Table: Create a table to allow one of the columns to contain unique values and the combination of another two are referencing to the column combination of another table


14. Write a SQL statement to create a table employees, including employee_id, first_name, last_name, email, phone_number hire_date, job_id, salary, commission, manager_id and department_id and make sure that, the employee_id column did not contain any duplicate values at the time of insertion and the foreign key columns combined by department_id and manager_id columns contain only those unique combination values, which combinations exist in the departments table.

Assume the structure of departments table below.


Indexes:
    "departments_pkey" PRIMARY KEY, btree (department_id, manager_id)

Sample Solution:

Code:

CREATE TABLE IF NOT EXISTS employees ( 
EMPLOYEE_ID decimal(6,0) NOT NULL PRIMARY KEY, 
FIRST_NAME varchar(20) DEFAULT NULL, 
LAST_NAME varchar(25) NOT NULL, 
EMAIL varchar(25) NOT NULL, 
PHONE_NUMBER varchar(20) DEFAULT NULL, 
HIRE_DATE date NOT NULL, 
JOB_ID varchar(10) NOT NULL, 
SALARY decimal(8,2) DEFAULT NULL, 
COMMISSION_PCT decimal(2,2) DEFAULT NULL, 
MANAGER_ID decimal(6,0) DEFAULT NULL, 
DEPARTMENT_ID decimal(4,0) DEFAULT NULL, 
FOREIGN KEY(DEPARTMENT_ID,MANAGER_ID) 
REFERENCES  departments(DEPARTMENT_ID,MANAGER_ID)
);

Output:

postgres=# CREATE TABLE IF NOT EXISTS employees (
postgres(# EMPLOYEE_ID decimal(6,0) NOT NULL PRIMARY KEY,
postgres(# FIRST_NAME varchar(20) DEFAULT NULL,
postgres(# LAST_NAME varchar(25) NOT NULL,
postgres(# EMAIL varchar(25) NOT NULL,
postgres(# PHONE_NUMBER varchar(20) DEFAULT NULL,
postgres(# HIRE_DATE date NOT NULL,
postgres(# JOB_ID varchar(10) NOT NULL,
postgres(# SALARY decimal(8,2) DEFAULT NULL,
postgres(# COMMISSION_PCT decimal(2,2) DEFAULT NULL,
postgres(# MANAGER_ID decimal(6,0) DEFAULT NULL,
postgres(# DEPARTMENT_ID decimal(4,0) DEFAULT NULL,
postgres(# FOREIGN KEY(DEPARTMENT_ID,MANAGER_ID)
postgres(# REFERENCES  departments(DEPARTMENT_ID,MANAGER_ID));
CREATE TABLE

Here is the command to see the structure of the created table :

postgres=# \d employees
                       Table "public.employees"
     Column     |         Type          |            Modifiers
----------------+-----------------------+---------------------------------
 employee_id    | numeric(6,0)          | not null
 first_name     | character varying(20) | default NULL::character varying
 last_name      | character varying(25) | not null
 email          | character varying(25) | not null
 phone_number   | character varying(20) | default NULL::character varying
 hire_date      | date                  | not null
 job_id         | character varying(10) | not null
 salary         | numeric(8,2)          | default NULL::numeric
 commission_pct | numeric(2,2)          | default NULL::numeric
 manager_id     | numeric(6,0)          | default NULL::numeric
 department_id  | numeric(4,0)          | default NULL::numeric
Indexes:
    "employees_pkey" PRIMARY KEY, btree (employee_id)
Foreign-key constraints:
    "employees_department_id_fkey" FOREIGN KEY (department_id, manager_id) REFERENCES departments(department_id, manager_id)

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Previous: Write a SQL statement to create a table job_history, including employee_id, start_date, end_date, job_id and department_id and make sure that, the employee_id column does not contain any duplicate values at the time of insertion and the foreign key column job_id contain only those values which exist in the jobs table.
Next: Write a sql statement to create a table employees, including columns employee_id, first_name, last_name, email, phone_number hire_date, job_id, salary, commission, manager_id and department_id and make sure that, the employee_id column does not contain any duplicate value at the time of insertion, and the foreign key column department_id, reference by the column department_id of departments table, can contain only those values which exist in the departments table and another foreign key column job_id, referenced by the column job_id of jobs table, can contain only those values which exist in the jobs table.

What is the difficulty level of this exercise?



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