The CEO ________ that Tony was appointed as the manager of the marketing department in today's meeting. A、announcedB、declinedC、attended
The CEO ________ that Tony was appointed as the manager of the marketing department in today's meeting.
A、announced
B、declined
C、attended
相关考题：
You created a view called EMP_DEPT_VU that contains three columns from the EMPLOYEES and DEPARTMENTS tables:EMPLOYEE_ID, EMPLOYEE_NAME AND DEPARTMENT_NAME.The DEPARTMENT_ID column of the EMPLOYEES table is the foreign key to the primary keyDEPARTMENT_ID column of the DEPARTMENTS table.You want to modify the view by adding a fourth column, MANAGER_ID of NUMBER data type from the EMPLOYEES tables.How can you accomplish this task? （）A. ALTER VIEW EMP_dept_vu (ADD manger_id NUMBER);B. MODIFY VIEW EMP_dept_vu (ADD manger_id NUMBER);C. ALTER VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employee e, departments d WHERE e.department _ id = d.department_id;D. MODIFY VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employees e, departments d WHERE e.department _ id = d.department_id;E. CREATE OR REPLACE VIEW emp_dept_vu AS SELECT employee_id, employee_name, department_name, manager_id FROM employees e, departments d WHERE e.department _ id = d.department_id;F. You must remove the existing view first, and then run the CREATE VIEW command with a new column list to modify a view.
Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEESColumn name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyEMP_NAME VARCHAR2 (30)JOB_ID VARCHAR2 (20)SALARY NUMBERMGR_ID NUMBER References EMPLOYEE_ID COLUMNDEPARTMENT ID NUMBER Foreign key to DEPARTMENT ID column of the DEPARTMENTS tableDEPARTMENTSColumn name Data type RemarksDEPARTMENT_ID NUMBER NOT NULL, Primary Key DEPARTMENT_NAME VARCHAR2(30)MGR_ID NUMBER References MGR_ID column of the EMPLOYEES tableEvaluate this SQL statement:SELECT employee_id, e.department_id, department_name, salaryFROM employees e, departments dWHERE e. department_id = d.department_id;Which SQL statement is equivalent to the above SQL statement? （）A. SELECT employee_id, department_id, department_name, salary FROM employees WHERE department_id IN (SELECT department_id FROM departments);B. SELECT employee_id, department_id, department_name, salary FROM employees NATURAL JOIN departments;C. SELECT employee_id, d.department_id, department_name, salary FROM employees e JOIN departments d ON e.department _ id = d. department_id;D. SELECT employee_id, department_id, department_name, Salary FROM employees JOIN departments USING (e.department_id, d.department_id);
Youexecutethefollowingblockofcode：SQLBEGINDBMS_RESOURCE_MANAGER.CREATE_SIMPLE_PLAN（SIMPLE_PLAN=’DEPARTMENTS’，CONSUMER_GROUP1=’PAYROLLL’，GROUP1_CPU=50，CONSUMER_GROUP2=’SALES’，GROUP2_CPU=25，CONSUMER_GROUP3=’MARKETING’，GROUP3_CPU=25）；END；SQL/Whatisaprerequisiteforusingthesimpleresourceplancreatedbyexecutingtheabovecode？（）A.Youmustassignuserstoconsumergroups.B.Youmustgranttheswitchprivilegetotheusers.C.Youmustcreatearesourceplandirective.D.Youmustspecifythecomplexresourceplan.
ExhibitExamine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their manager‘s last names and their department names. Which query would you use?（）A. SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN department d ON (e.department_id = d.department_id);B. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.managaer_id = m.employee_id) LEFT OUTER JOIN department d ON (e.department_id = d.department_id);C. SELECT e.last_name, m.last_name, department_name FROM employees e RIGT OUTER JOIN employees m on ( e.manager_id = m.employee_id) FULL OUTER JOIN department d ON (e.department_id = d.department_id);D. SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGT OUTER JOIN department d ON (e.department_id = d.department_id);E. SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id)F. SELECT last_name, manager_id, department_name FROM employees e JOIN department d ON (e.department_id = d.department_id);
Examine the structure of the EMPLOYEES and DEPARTMENTS tables:EMPLOYEESEMPLOYEE_ID NUMBERDEPARTMENT_ID NUMBERMANAGER_ID NUMBERLAST_NAME VARCHAR2(25)DEPARTMENTSDEPARTMENT_ID NUMBERMANAGER_ID NUMBERDEPARTMENT_NAME VARCHAR2(35)LOCATION_ID NUMBERYou want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?（）A. SELECT last_name, department_name, location_id FROM employees , department ;B. SELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;C. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;D. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
阅读以下说明和Java代码，将应填入(n)处的字句写在对应栏内。[说明]本程序的功能是给公司的员工Tom，Jack，Green增加薪水。三人的职位分别是programmer， Manager，CEO。程序由6个类组成：WorkerTest是主类，programmer，Manager，CEO三个类，薪水增加的规则是 programmer的涨幅是5%；Manager的是10%；CEO也是Manager，但是它除了有Manager的涨幅，还有1000元的bonus。接口SalaryRaise提供了一个增加薪水的方法raise()。[java程序]public class WorkerTest {public WorkerTest( ) {}public static void main( String[] args) {Programmer programmer = new Programmer( "Tom" ,3000);Manager manager = new Manager( "Jack" ,4000);CEO ceo = new CEO( "Green" ,4000);Worker [] worker = new Worker[3];programmer, raise( );manager, raise( );ceo. raise( );worker[0] = programmer;worker [1] = manager;worker[2] = ceo;for ( int i = 0 ;i ＜ worker, length; i + + ) {System. out. prinfln (" Name:" + worker [i]. getName ( ) +" \ tSalary:" + worker [i]. getSalary ());public interface SalaryRaise { void raise( ); }public class Worker {public String name;public double (1);public Worker( ) {}public String getName( ) {return name;}public void setName( String name) {this. name = name;}public double getSalary( ) {return salary;}public void setSalary(double salary) { this. salary = salary; }}public class Programmer extends Worker implements (2) {public Programmer( ) {}public void raise( ) {double pets=0.05;double sala = this. getSalary( ) * (1 + pers);this. setSalary (sala);public Programmer( Siring name, double salary) tthis. name = name;this. salary = salary;public class Manager extends (3) implements SalaryRaise {public Manager( ) { }public Manager(String name, double salary) {this. name = name;this. salary = salary;}public void raise( ) {double pets = 0.1;double sala = this. getSalary() * (1 + pers);this. setSalary(sala);}}public class CEO extends Manager implements SalaryRaise {public CEO() {}public CEO( String name,double salary) {this. name = name;this. salary = salary;}public void raise( ) {double bonus = 1000;(4);double sala = this. getSalary( );(5);this. setSalary(sala);}}
Examine the data in the EMPLOYEES and DEPARTMENTS tables.EMPLOYEESLAST_NAME DEPARTMENT_ID SALARYGetz 10 3000Davis 20 1500Bill 20 2200Davis 30 5000Kochhar 5000DEPARTMENTSDEPARTMENT_ID DEPARTMENT_NAME10 Sales20 Marketing30 Accounts40 AdministrationYou want to retrieve all employees, whether or not they have matching departments in the departments table.Which query would you use?（）A. SELECT last_name, department_name FROM employees , departments(+);B. SELECT last_name, department_name FROM employees JOIN departments(+);C. SELECT last_name, department_name ON (e. department_ id = d. departments_id); FROM employees(+) e JOIN departments dD. SELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);E. SELECT last_name, department_name FROM employees(+) , departments ON (e. department _ id = d. department _id);F. SELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e. department _ id = d. department _id);
Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their managers‘ last names and their department names. Which query would you use?（）A.SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN departments d ON (e.department_id = d.department_id);B.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);C.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);D.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);E.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);F.SELECT last_name, manager_id, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id) ;
This is the first time that a woman has been______to the post.A．grantedB．praisedC．pointedD．appointed
给出下面不完整的类代码，则横线处的语句应该为( )。 class Person { String name，department； int age； public Person (Strings) {name=s；} public Person (String s，intA．{name=s；age=a;} public Person (String n，String d，intA){ __________ department=d； } }A)Person (n，A)；B．this (Person(n，A))；C．this(n，A)；D．this(name，age)；