Question

Given a table Employees(EmpID, ManagerID, Salary), which SQL statement correctly returns the EmpID of every employee who earns strictly more than their direct manager?

Q: Answer-Given a table Employees(EmpID, ManagerID, Salary), which SQL statement correctly returns the EmpID o...

Each employees manager is the row whose EmpID equals the employees ManagerID, so the correct self-join condition is e.ManagerID m.EmpID, followed by comparing e.Salary m.Salary. Option A joins employees sharing the same manager peers, not the manager itself. Options B and C reverse the join direction, matching an employee to their subordinates rather than their manager. Option E compares against the average salary of an employees peer group, not the managers actual salary.

A SELECT e.EmpID FROM Employees e, Employees m WHERE e.ManagerID = m.ManagerID AND e.Salary > m.Salary;

B SELECT e.EmpID FROM Employees e LEFT JOIN Employees m ON e.EmpID = m.ManagerID WHERE e.Salary > m.Salary;

C SELECT e.EmpID FROM Employees e JOIN Employees m ON e.EmpID = m.ManagerID WHERE e.Salary > m.Salary;

D SELECT e.EmpID FROM Employees e JOIN Employees m ON e.ManagerID = m.EmpID WHERE e.Salary > m.Salary;

E SELECT e.EmpID FROM Employees e WHERE e.Salary > (SELECT AVG(Salary) FROM Employees m WHERE m.ManagerID = e.ManagerID);

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Given a table Employees(EmpID, ManagerID, Salary), which SQL statement correctly returns the EmpID of every employee who earns strictly more than their direct manager?

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