分析下面的JavaScript代码段,输出的结果是( )。emp=new Array(3);emp[0]=0;emp[1]=1;emp[2]=2;emp[3]=3;document.write(emp.length); A.2B.3C.4D.5
分析下面的JavaScript代码段,输出的结果是( )。emp=new Array(3);emp[0]=0;emp[1]=1;emp[2]=2;emp[3]=3;document.write(emp.length);
A.2
B.3
C.4
D.5
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以下插入记录正确的( )A.insert into emp(ename,hiredate,sal) values (value1,value2,value3);B.insert into emp (ename,sal)values(value1,value2,value3);C.insert into emp (ename)values(value1,value2,value3);D.insert into emp (ename,hiredate,sal)values(value1,value2);
Which is a valid CREATE TABLE statement? () A. CREATE TABLE EMP9$# AS (empid number(2));B. CREATE TABLE EMP*123 AS (empid number(2));C. CREATE TABLE PACKAGE AS (packid number(2));D. CREATE TABLE 1EMP_TEST AS (empid number(2));
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();}
● 假设员工关系EMP(员工号,姓名,部门,部门电话,部门负责人,家庭住址,家庭成员,成员关系)如下表所示。如果一个部门可以有多名员工,一个员工可以有多个家庭成员,那么关系EMP属于 (52) ,且 (53) 问题;为了解决这一问题,应该将员工关系EMP分解为 (54) 。(52)A. 1NFB. 2NFC. 3NFD. BCNF(53)A. 无冗余、无插入异常和删除异常B. 无冗余,但存在插入异常和删除异常C. 存在冗余,但不存在修改操作的不一致D. 存在冗余、修改操作的不一致,以及插入异常和删除异常(54)A. EMP1(员工号,姓名,家庭住址)EMP2(部门,部门电话,部门负责人)EMP3(员工号,家庭成员,成员关系)B. EMP1(员工号,姓名,部门,家庭住址)EMP2(部门,部门电话,部门负责人)EMP3(员工号,家庭成员,成员关系)C. EMP1(员工号,姓名,家庭住址)EMP2(部门,部门电话,部门负责人,家庭成员,成员关系)D. EMP1(员工号,姓名,部门,部门电话,部门负责人,家庭住址)EMP2(员工号,家庭住址,家庭成员,成员关系)
● 假设员工关系 EMP(员工号,姓名,性别,部门,部门电话,部门负责人,家庭住址,家庭成员,成员关系)如下表所示。如果一个部门只能有一部电话和一位负责人,一个员工可以有多个家庭成员,那么关系EMP属于 (31) ,且 (32) 问题;为了解决这一问题,应该将员工关系EMP分解为 (33) 。(31)A. 1NFB. 2NFC. 3NFD. BCNF(32)A. 无冗余、无插入异常和删除异常B. 无冗余,但存在插入异常和删除异常C. 存在冗余,但不存在修改操作的不一致D. 存在冗余、修改操作的不一致,以及插入异常和删除异常(33)A. EMP1(员工号,姓名,性别,家庭住址)EMP2(部门,部门电话,部门负责人)EMP3(员工号,家庭成员,成员关系)B. EMP1(员工号,姓名,性别,部门,家庭住址)EMP2(部门,部门电话,部门负责人)EMP3(员工号,家庭成员,成员关系)C.EMP1(员工号,姓名,性别,家庭住址)EMP2(部门,部门电话,部门负责人,家庭成员,成员关系)D. EMP1(员工号,姓名,性别,部门,部门电话,部门负责人,家庭住址)EMP2(员工号,家庭住址,家庭成员,成员关系)
( )A.π2,4 (σ8='开发部'(EMP×DEP))B.π2,4 (σ1=9(EMP σ2='开发部'(DEP)))C.π2,3 (EMP×σ2='开发部'(DEP))D.π2,3 (π1,2,4,6(EMP) σ2='开发部'(DEP))
Select emp_id,emp_name,sex,title,wage from employee order by emp_name句子得到的结果集是按()排序。A.emp_idB.emp_nameC.sexD.wage
分析下面的JavaScript代码段,输出的结果是( )。emp=new Array(5);emp[1]=1;emp[2]=2;document.write(emp.length); A.2B.3C.4D.5
emp表是雇员信息表,以下哪个变量可以存放emp表中的一条记录() A.v_record emp%type;B.v_record emp%recordtype;C.v_record emp%record_type;D.v_record emp%rowtype;E.v_record emp%row_type;
有如下三步操作,并且均成功执行,请选择结果()delete from emp;alter table emp add(addr varchar2(40));rollback; A. emp表中数据处于回滚段中B. emp表中的数据被永久删除C. emp表中的数据没有变化D. 原来的emp表中的数据被添加到新emp表中E. 以上所述都不正确
分析下面的JavaScript代码段,输出的结果是()。 emp=new Array(5); emp[1]=1 emp[2]=2 document.write(emp.length);A、2B、3C、4D、5
在JavaScript中,下列定义数组的方法正确的是()。A、emp=newArray[3];B、emp=newArray(3);C、emp=new()array;D、emp=newArray[];
Select emp_id,emp_name,sex,title,wage from employee order by emp_name句子得到的结果集是按()排序。A、emp_idB、emp_nameC、sexD、wage
分析下面的JavaScript代码段,输出的结果是() emp = new Array(5); emp[1]=1; emp[2]=2; document.write(emp.length);A、2B、3C、4D、5
有如下三步操作,并且均成功执行,请选择结果() delete from emp; alter table emp add(addr varchar2(40)); rollback;A、emp表中数据处于回滚段中B、emp表中的数据被永久删除C、emp表中的数据没有变化D、原来的emp表中的数据被添加到新emp表中E、以上所述都不正确
单选题分析下面的JavaScript代码段,输出的结果是() emp = new Array(5); emp[1]=1; emp[2]=2; document.write(emp.length);A2B3C4D5
单选题查询出EMP表中COMM字段为空的记录()Aselect * from emp where comm='';Bselect * from emp where comm=null;Cselect * from emp where nvl(comm)=0;Dselect * from emp where comm is null;Eselect * from emp where nvl(comm,0)=0;
单选题分析下面的JavaScript代码段,输出的结果是()。 emp=new Array(5); emp[1]=1 emp[2]=2 document.write(emp.length);A2B3C4D5
单选题Which SQL statement defines a FOREIGN KEY constraint on the DEPTNO column of the EMP table?()ACREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);BCREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));CCREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));DCREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
单选题Which is a valid CREATE TABLE statement?()ACREATE TABLE EMP9$# AS (empid number(2));BCREATE TABLE EMP*123 AS (empid number(2));CCREATE TABLE PACKAGE AS (packid number(2));DCREATE TABLE 1EMP_TEST AS (empid number(2));
单选题emp表是雇员信息表,以下哪个变量可以存放emp表中的一条记录()Av_record emp%type;Bv_record emp%recordtype;Cv_record emp%record_type;Dv_record emp%rowtype;Ev_record emp%row_type;
单选题Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE You issue these statements: CREATE table new_emp ( employee_id NUMBER, name VARCHAR2(30)); INSERT INTO new_emp SELECT employee_id , last_name from employees; Savepoint s1; UPDATE new_emp set name = UPPER(name); Savepoint s2; Delete from new_emp; Rollback to s2; Delete from new_emp where employee_id =180; UPDATE new_emp set name = 'James'; Rollback to s2; UPDATE new_emp set name = 'James' WHERE employee_id =180; Rollback; At the end of this transaction, what is true?()AYou have no rows in the table.BYou have an employee with the name of James.CYou cannot roll back to the same savepoint more than once.DYour last update fails to update any rows because employee ID 180 was already deleted.
单选题Which SQL statement defines the FOREIGN KEY constraint on the DEPTNO column of the EMP table?()ACREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) NOT NULL CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);BCREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));CCREATE TABLE EMP (empno NUMBER(4) ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));DCREATE TABLE EMP (empno NUMBER(4), ename VARCNAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp deptno fk REFERENCES dept (deptno));
单选题Management has asked you to calculate the value 12*salary* commission_pct for all the employees in the EMP table. The EMP table contains these columns: LAST NAME VARCNAR2(35) NOT NULL SALARY NUMBER(9,2) NOT NULL COMMISION_PCT NUMBER(4,2) Which statement ensures that a value is displayed in the calculated columns for all employees?()ASELECT last_name, 12*salary* commission_pct FROM emp;BSELECT last_name, 12*salary* (commission_pct,0) FROM emp;CSELECT last_name, 12*salary*(nvl(commission_pct,0)) FROM emp;DSELECT last_name, 12*salary*(decode(commission_pct,0)) FROM emp;
单选题您已创建一个名为 CalcSalary,将确定 Certkiller.com 员工的薪酬类的责任。CalcSalary 类包括员工的薪酬递增和递减的方法。下面的代码包含在 CalcSalary 类中:()public class CalcSalary {// for promotionspublic static bool IncrementSalary (Employee Emp, double Amount){if (Emp.Status == QuarterlyReview.AboveGoals)Emp.Salary += Amount;return true;Apublic delegate bool Salary (Employee Emp, double Amount);Bpublic bool Salary (Employee Emp, double Amount);Cpublic event bool Salary (Employee Emp, double Amount);Dpublic delegate void Salary (Employee Emp, double Amount);