The EMP table contains these columns:LAST NAME VARCHAR2(25)SALARY NUMBER(6,2)DEPARTMENT_ID NUMBER(6)You need to display the employees who have not been assigned to any department. You write the SELECT statement:SELECT LAST_NAME, SALARY, DEPARTMENT_ID FROM

Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary KeyFIRST_NAME VARCHAR2(25)LAST_NAME VARCHAR2(25)DEPARTMENT_ID NUMBERSALARY NUMBERWhat is the correct syntax for an inline view? （）

A. SELECT a.last_name, a.salary, a.department_id, b.maxsal FROM employees a, (SELECT department_id, max(salary)maxsal FROM employees GROUP BY department_id) b WHERE a.department_id = b.department_id AND a.salary < b.maxsal;

B. SELECT a.last name, a.salary, a.department_id FROM employees a WHERE a.department_id IN (SELECT department_id FROM employees b GROUP BY department_id having salary = (SELECT max(salary) from employees))

C. SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE a.salary = (SELECT max(salary) FROM employees b WHERE a.department _ id = b.department _ id);

D. SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE (a.department_id, a.salary) IN (SELECT department_id, a.salary) IN (SELECT department_id max(salary) FROM employees b GROUP BY department_id ORDER BY department _ id);

阅读以下说明C++代码，将应填入(n)处的字句写在对应栏内。

[说明]

本程序实现了雇员信息管理功能，其中封装了雇员信息及其设置、修改、删除操作。已知当输入为“Smith 31 2960.0”时，程序的输出是：

姓名：Smith 年龄：31 工资：2960

姓名：Smith 年龄：31 工资：3500

姓名：Mary 年龄：23 工资：2500

[C++程序]

include ＜iostream.h＞

include ＜string.h＞

class employee{

char *name; //雇员姓名

short age; //年龄

float salary;//工资

public:

employee();

void set_name(char *);

void set_age(short a) {age=a;}

void set_salary(float s) {salary=s;}

(1);

~ employee(){delete[] name;}

};

employee::employee() { name="";

age=0;

salary=0.0;

void employee::set_name(char *n)

{ name=new char[strlen(n)+1];

(2) (name,n);

}

void employee::print()

{ cout＜＜"姓名":"＜＜name＜＜" 年龄:"＜＜agc＜＜" 工资:" ＜＜salary＜＜endl;

}

void main()

{ char *na;

short ag=0;

float sa=0;

(3);

na=new char[10];

cin＞＞na＞＞ag＞＞sa;

emp.set_name(na);

emp.set_age(ag);

emp.set_salary(sa);

emp.print();

(4) (3500.0);

emp.print();

(5);

emp.set_name("Mary");

emp.set_age(23);

emp.set_salary(2500.0);

emp.print();

}

Click the Exhibit button to examine the data of the EMPLOYEES table.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?（）

A.SELECT employee_id "Emp_id", emp_name "Employee", salary, employee_id "Mgr_id", emp_name "Manager" FROM employees WHERE salary > 4000;

B.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m WHERE e.mgr_id = m.mgr_id AND e.salary > 4000;

C.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m ON (e.mgr_id = m.employee_id) AND e.salary > 4000;

D.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id", m.emp_name "Manager" FROM employees e SELF JOIN employees m WHERE e.mgr_id = m.employee_id AND e.salary > 4000;

E.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id" m.emp_name "Manager" FROM employees e JOIN employees m USING (e.employee_id = m.employee_id) AND e.salary > 4000;

The EMPLOYEE tables has these columns:LAST_NAME VARCHAR2(35)SALARY NUMBER(8,2)COMMISSION_PCT NUMBER(5,2)You want to display the name and annual salary multiplied by the commission_pct for all employees. For records that have a NULL commission_pct, a zero must be displayed against the calculated column.Which SQL statement displays the desired results? （）

A. SELECT last_name, (salary * 12) * commission_pct FROM EMPLOYEES;

B. SELECT last_name, (salary * 12) * IFNULL(commission_pct, 0) FROM EMPLOYEES;

C. SELECT last_name, (salary * 12) * NVL2(commission_pct, 0) FROM EMPLOYEES;

D. SELECT last_name, (salary * 12) * NVL(commission_pct, 0) FROM EMPLOYEES;

The EMP table contains these columns:LAST_NAME VARCHAR2 (25)SALARY NUMBER (6,2)DEPARTMENT_ID NUMBER (6)You need to display the employees who have not been assigned to any department. You write the SELECT statement:SELECT LAST_NAME, SALARY, DEPARTMENT_IDFROM EMPWHERE DEPARTMENT_ID = NULL;What is true about this SQL statement ?（）

A.The SQL statement displays the desired results.

B.The column in the WHERE clause should be changed to display the desired results.

C.The operator in the WHERE clause should be changed to display the desired results.

D.The WHERE clause should be changed to use an outer join to display the desired results.