sql.js

Enter some SQL

DROP TABLE IF EXISTS employees;
CREATE TABLE employees( id          integer,  name    text,
                          designation text,     manager integer,
                          hired_on    date,     salary  integer,
                          commission  float,    dept    integer);

INSERT INTO employees VALUES (1,'JOHNSON','ADMIN',6,'12-17-1990',18000,NULL,4);
  INSERT INTO employees VALUES (2,'HARDING','MANAGER',9,'02-02-1998',52000,300,3);
  INSERT INTO employees VALUES (3,'TAFT','SALES I',2,'01-02-1996',25000,500,3);
  INSERT INTO employees VALUES (4,'HOOVER','SALES I',2,'04-02-1990',27000,NULL,3);
  INSERT INTO employees VALUES (5,'LINCOLN','TECH',6,'06-23-1994',22500,1400,4);
  INSERT INTO employees VALUES (6,'GARFIELD','MANAGER',9,'05-01-1993',54000,NULL,4);
  INSERT INTO employees VALUES (7,'POLK','TECH',6,'09-22-1997',25000,NULL,4);
  INSERT INTO employees VALUES (8,'GRANT','ENGINEER',10,'03-30-1997',32000,NULL,2);
  INSERT INTO employees VALUES (9,'JACKSON','CEO',NULL,'01-01-1990',75000,NULL,4);
  INSERT INTO employees VALUES (10,'FILLMORE','MANAGER',9,'08-09-1994',56000,NULL,2);
  INSERT INTO employees VALUES (11,'ADAMS','ENGINEER',10,'03-15-1996',34000,NULL,2);
  INSERT INTO employees VALUES (12,'WASHINGTON','ADMIN',6,'04-16-1998',18000,NULL,4);
  INSERT INTO employees VALUES (13,'MONROE','ENGINEER',10,'12-03-2000',30000,NULL,2);
  INSERT INTO employees VALUES (14,'ROOSEVELT','CPA',9,'10-12-1995',35000,NULL,1);

SELECT designation,COUNT(*) AS nbr, (AVG(salary)) AS avg_salary FROM employees GROUP BY designation ORDER BY avg_salary DESC;
SELECT name,hired_on FROM employees ORDER BY hired_on;

Load an SQLite database file:

Result:

Online SQL interpreter