Consider the following relational schema.
Employee(E_ID, E_Name, Dept_Name, Salary, Address)
Now write down the SQL for the following:
- Display E_Name who is taking maximum salary.
- Display all Dept_Name where number of Employees are less than 10.
- Display E_Name of employee whose address is either Mechi, Nepalgunj or Pokhara.
- Display E_ID and E_Name of employees whose name starts with “S” and Ends with “M”.
- Increase the salary of employees by 50% whose salary is less than 10000.
A: DBMS: SQL for the Table Employee (E_ID, E_Name, Dept_Name, Salary, Address) : 31
i. Display E_Name who is taking maximum salary.
SELECT E_Name
FROM Employee
WHERE Salary = (SELECT MAX(Salary) FROM Employee);
---- OR ----
SELECT E_Name FROM Employee
ORDER BY Salary DESC
LIMIT 1;
ii. Display all Dept_Name where number of Employees are less than 10.
SELECT Dept_Name, COUNT(E_ID)
FROM Employee
GROUP BY Dept_Name
HAVING COUNT(E_ID) < 10;
iii. Display E_Name of employee whose address is either Mechi, Nepalgunj or Pokhara.
SELECT E_Name
FROM Employee
WHERE Address IN ( "Mechi", "Nepalgunj", "Pokhara" );
---- OR ----
SELECT E_Name
FROM Employee
WHERE Address = "Mechi" OR Address = "Nepalgunj"
OR Address = "Pokhara" ;
iv. Display E_ID and E_Name of employees whose name starts with “S” and ends with “M”.
SELECT E_ID, E_Name
FROM Employee
WHERE E_Name LIKE "S%M";
v. Increase the salary of employees by 50% whose salary is less than 10000.
UPDATE Employee
SET Salary = Salary * 1.5
WHERE Salary < 10000;