我的实验室任务有些问题:
当我的程序试图提示用户输入时,程序在同一行输出两个问题,只接受第二个问题的输入。
我的程序输出:
请输入第二名员工的姓名:请输入第二名员工的姓名:1
(它们出现在同一行而不是单独的行)
数组输出也是如下:
0.00.00.00.00.00.00.00.00.00.0
而不是像这样:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
我不太确定如何解决这两个问题,我们将不胜感激!
这是我的代码:
Employee.java
//import java.util.*;
public class Employee
{
private String empName;
private int empNumber;
private String empAddress;
private double empSalary;
private double[] empBonus=new double[10];
public Employee(){}
public Employee(String empName_, int empNumber_, String empAddress_, double empSalary_, double[] empBonus_)
{
this.empName=empName_;
this.empNumber=empNumber_;
this.empAddress=empAddress_;
this.empSalary=empSalary_;
this.empBonus=empBonus_;
}
public String getName()
{
return this.empName;
}
public int getEmployeeNumber()
{
return this.empNumber;
}
public String getAddress()
{
return this.empAddress;
}
public double getSalary()
{
return this.empSalary;
}
public String changeAddress(String chAddress)
{
return empAddress=chAddress;
}
public double changeSalary(double chSalary)
{
return empSalary=chSalary;
}
public String addBonus(double[] empBonus)
{
String arrayBonus=new String("");
for(int i=0; i<empBonus.length;i++)
{
arrayBonus+=empBonus[i];
}
return arrayBonus;
}
public String toString()
{
return ("\nEmployee's name: "+empName+"\nEmployee's Number: "+empNumber+"\nEmployee's address: "+empAddress+
"\nEmployee's original salary: "+empSalary+ "\nEmployee's bonuses: "+addBonus(empBonus)+"\n");
}
}
EmployeeTester.java
import java.util.*;
public class EmployeeTester
{
public static void main(String[] args)
{
Scanner in1=new Scanner(System.in);
Scanner in2=new Scanner(System.in);
Scanner in3=new Scanner(System.in);
Employee emp1;
Employee emp2;
emp1=read_input("first", in1, in2, in3);
emp2=read_input("second", in1, in2, in3);
System.out.println(emp1.toString());
System.out.println(emp2.toString());
}
public static Employee read_input(String msg, Scanner scan1, Scanner scan2, Scanner scan3)
{
String name, address;
int num;
double salary;
double[] bonus=new double[10];
System.out.print("\nPlease enter the name of the "+msg+" employee:");
name=scan1.nextLine();
System.out.print("Please enter the number of the "+msg+" employee:");
num=scan2.nextInt();
System.out.print("Please enter the address of the "+msg+" employee:");
address=scan1.nextLine();
System.out.print("Please enter the salary of the "+msg+" employee:");
salary=scan3.nextDouble();
System.out.print("Please add a bonus for the "+msg+" employee:");
bonus[0]=scan3.nextDouble();
System.out.print("Add more bonuses to the "+msg+"employee? (y/n) \nNote: Enter 0.0 if you would like to terminate adding more bonuses: ");
if(scan1.next().startsWith("y"))
{
for(int i=1; i<bonus.length;i++)
{
System.out.print("Continue entering a bonus to "+msg+" employee:");
bonus[i]=scan3.nextDouble();
if(bonus[i]==0.0 || i==bonus.length)
{
break;
}
}
}
return new Employee(name, num, address, salary, bonus);
}
}
答案 0 :(得分:2)
对于您的第一个问题,只需将Scanners
移至read_input
方法内,以便每次都重新开始。
public static void main(String[] args) {
Employee emp1;
Employee emp2;
emp1 = read_input("first");
emp2 = read_input("second");
System.out.println(emp1.toString());
System.out.println(emp2.toString());
}
public static Employee read_input(String msg) {
Scanner scan1 = new Scanner(System.in);
Scanner scan2 = new Scanner(System.in);
Scanner scan3 = new Scanner(System.in);
...
对于第二个问题,在构建输出字符串的addBonus
方法中,不添加任何空格或逗号。如果对这种类型的循环连接使用StringBuilder
而不是重复创建新的字符串对象,那么效率会更高。
public String addBonus(double[] empBonus)
{
StringBuilder arrayBonus = new StringBuilder();
for(int i=0; i<empBonus.length;i++)
{
arrayBonus.append(empBonus[i] + ", ");
}
return arrayBonus.toString();
}
答案 1 :(得分:0)
您不需要3种不同的扫描仪,一种是扫描仪。
对于打印部件,您需要println()
或使用'\n'
作为新行。
E.g:
System.out.println("Please enter the name of the "+msg+" employee:");
System.out.print("Please enter the name of the "+msg+" employee:\n");
在第一种情况下,您正在调用一个函数(println()
而不是print()
),该函数会自动将新行添加到outputes文本中。在第二种情况下,您正在创建字符串的新行部分(您已经在第一行的开头使用了它)
数组输出使用0.0
,因为值是浮点值(在您的情况下为double
),默认情况下会打印小数部分。要仅打印整数部分,您需要将值转换为整数或使用double的格式。
铸造:
double a = 0.0;
System.out.print("'a' as integer: " + ((int)a));
格式:
System.out.format("Here is a number: %.0f", a);
.
和f
之间的数字指定要输出的小数位数。