Find the IDs and names of the employees whose salaries are at least 10% higher than the average salary of all employees

Question

I need to find the IDs and names of the employees whose salaries are at least 10% higher than the average salary of all employees. I can't get the code to work with the 10% higher than the average. I know how to get the average, but when I run my particular code, I get all names and know that it's not running correctly.

Here is my query:

Select employees.eid, 
       CONCAT(efirstName, 
       ' ', eLastName) AS Name 
from employees 
GROUP BY employees.salary 
HAVING employees.salary > avg(salary * .10); 

I have also tried:

Select employees.eid, 
       CONCAT(eFirstName, 
       ' ', 
       eLastName AS Name 
FROM employees 
GROUP by eLastName 
HAVING avg(salary) > avg(salary * .10);

Answer 1

Using a simple subquery should help with this:

SELECT e.eid, CONCAT(efirstname, ' ', e.elastname) name
FROM employees e
WHERE e.salary > (SELECT AVG(salary) FROM employees) * 1.10

Answer 2

Can u try this:

    SELECT e.eid, concat(e.eFirstName, ' ', e.LastName) as name
    FROM employees e
    WHERE e.salary > AVG(e.salary) * 0.1;