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【OCP-12c】CUUG最新考試原題整理及答案(071-10)

ast play lec select employee ocp tput for ans

10、(5-6) choose the best answer:
Examine the structure of the EMPLOYEES table:
There is a parent/child relationship between EMPLOYEE_ID and MANAGER_ID.
You want to display the name, joining date, and manager for all the employees.
Newly hired employees are yet to be assigned a department or a manager. For them, ‘No Manager‘ should be displayed in the MANAGER column.

Which SQL query gets the required output?
A) SELECT e.last_name, e.hire_date, NVL(m.last_name, ‘No Manager‘) Manager
FROM employees e RIGHT OUTER JOIN employees m
ON (e. manager_id = m.employee_id);

B) SELECT e.last_name, e.hire_date, NVL(m.last_name, ‘No Manager‘) Manager
FROM employees e JOIN employees m

ON (e.manager_id = m.employee_id);

C) SELECT e.last_name, e.hire_date, NVL(m.last_name, ‘No Manager‘) Manager
FROM employees e LEFT OUTER JOIN employees m
ON (e.manager_id = m.employee_id);

D) SELECT e.last_name, e.hire_date, NVL(m.last_name, ‘No Manager‘) Manager
FROM employees e NATURAL JOIN employees m

ON (e.manager_id = m.employee_id);

Answer:C
(解析:因為員工 king 是沒有經理的,但是也要顯示出來,因為是在經理號這邊缺少數據,所以這裏要用
左外連接。註意左右的區別:
SELECT e.last_name, e.hire_date, NVL(m.last_name, ‘No Manager‘) Manager
FROM employees e , employees m
WHERE e.manager_id = m.employee_id(+);

【OCP-12c】CUUG最新考試原題整理及答案(071-10)