COUNT with RANK mysql

Question

From my database I want to get the lastname of employee, the number of employees who earn higher than him, and then rank the employee's salary in descending order; with the highest earning employee as zero. Here is how I have done it:

SELECT 0+COUNT(b.salary) rank
, a.lname
, a.salary
, COUNT(*) AS employee_count
FROM employee a LEFT OUTER JOIN employee b
ON a.salary<b.salary
GROUP BY a.salary, a.lname
ORDER BY salary DESC

And the result is shown like this:

So far so good. Except, Borg shouldn't be even there. Because no one is earning more than him. Now this happened because of using LEFT OUTER JOIN. I tried using INNER JOIN but it gives syntax error. So the question is how can I make INNER JOIN work with this?

Answer 1

If you just want to eliminate the top value:

SELECT *
FROM (    
    SELECT 0+COUNT(b.salary) rank
        , a.lname
        , a.salary
        , COUNT(*) AS employee_count
    FROM employee a 
    LEFT OUTER JOIN employee b
        ON a.salary < b.salary
    GROUP BY a.salary, a.lname
    ORDER BY salary DESC
) T1
WHERE T1.rank > 0