1Z0-007 · Question #13
Examine the data of the EMPLOYEES table. EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID) Which statement lists the ID, name, and salary of the…
The correct answer is C. SELECT e.employee_id "Emp_id", e.emp_name "Employee". This statement lists the ID, name, and salary of the employee, and the ID and name of the employee's manager, for all the employees who have a manager and earn more than 4000 Incorrect Answers A: This statement does not check does employee have a manager or not, so it will not…
Question
Examine the data of the EMPLOYEES table. EMPLOYEES (EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID) Which statement lists the ID, name, and salary of the employee, and the ID and name of the employee's manager, for all the employees who have a manager and earn more than 4000?
Exhibit
Options
- ASELECT employee_id "Emp_id", emp_name "Employee",
- BSELECT e.employee_id "Emp_id", e.emp_name "Employee",
- CSELECT e.employee_id "Emp_id", e.emp_name "Employee",
- DSELECT e.employee_id "Emp_id", e.emp_name "Employee",
- ESELECT e.employee_id "Emp_id", e.emp_name "Employee",
How the community answered
(34 responses)- A6% (2)
- B3% (1)
- C76% (26)
- D12% (4)
- E3% (1)
Explanation
This statement lists the ID, name, and salary of the employee, and the ID and name of the employee's manager, for all the employees who have a manager and earn more than 4000 Incorrect Answers A: This statement does not check does employee have a manager or not, so it will not provide B: Usage of "e.mgr_id = m.mgr_id" condition is wrong to achieve required result. D: This statement uses "m.mgr_id" to show manager's manager, not employ's manager. E: Usage of "WHERE E.employee_id = m.employee_id" condition is wrong to achieve required result.
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