本文主要向大家介绍了Oracle数据库之【OCP-12c】CUUG最新考试原题整理及答案（071-10），通过具体的内容向大家展现，希望对大家学习Oracle数据库有所帮助。

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：因为员工 king 是没有经理的，但是也要显示出来，因为是在经理号这边缺少数据，所以这里要用
左外连接。注意左右的区别：
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(+);
）

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