Question

我正在尝试使用扫描程序类的用户输入值firstname，middlename，lastname，age，但是下面的程序只接受firstname，middlename，lastname而不是age的值。

public void inputEmployeeDetails(){
     Scanner  scanner = new Scanner(System.in);
                      System.out.println("Enter the firstname ");
                      firstname= scanner.nextLine();
                     System.out.println("Enter the middlename ");
                      middlename= scanner.nextLine();
                      System.out.println("Enter the lastname");
                      lastname= scanner.nextLine();
                     System.out.println("Enter the age"); 
                     age= scanner.nextInt();
     }

我需要通过用户提示一个接一个地获取所有值。根据输入的数据，我显示员工详细信息。有人能让我知道我错过了什么。

我还想从用户那里获取多个输入。如果我在下面，请告诉我

 System.out.println("Do you like to fetch more records press "Yes" or "No");
       String input=scanner.nextLine();
       if(input="Yes")
       inputEmployeeDetails();

Answer 1

年龄变量采用您按下的下一个ENTER键，因此不存储任何内容。由于ENTER的数据类型与age变量的数据类型不同，因此您的代码正在获取InputMismatchException。您需要检查下一个值是否为int，然后将其分配给变量。请参考以下一次：

import java.util.Scanner;
public class Test{
     public static void main(String []args){
     String firstname, middlename, lastname;
     int age;
     Scanner  scanner = new Scanner(System.in);
     System.out.println("Enter the firstname ");
     firstname= scanner.nextLine();
     System.out.println("Enter the middlename ");
     middlename= scanner.nextLine();
     System.out.println("Enter the lastname");
     lastname= scanner.nextLine();
     System.out.println("Enter the age"); 
     if(scanner.hasNextInt()) {
        age= scanner.nextInt();
     }else
        age = 0;
     System.out.println(firstname + " " + middlename + " " + lastname + " " + age); 
    }
}

Answer 2

检查此程序：

String firstname,middlename,lastname,choice;
        int age, flag;
        Scanner  scanner = new Scanner(System.in);
        do{
             choice=null;

         System.out.println("Enter the firstname ");
         firstname= scanner.nextLine();
        System.out.println("Enter the middlename ");
         middlename= scanner.nextLine();
         System.out.println("Enter the lastname");
         lastname= scanner.nextLine();
        System.out.println("Enter the age");

        flag=0;age=0;

        while(flag==0){
        try{
        age= scanner.nextInt();
        flag=1;
        }
        catch(Exception e){
            scanner.nextLine();
            System.out.println("Wrong entry, please enter digits only:");
        }
        }

        System.out.println(firstname+" "+middlename+" "+lastname+" "+age);

        System.out.println("Do you like to fetch more records press Yes or No");
        scanner.nextLine();   

                   choice=scanner.nextLine();
        }while(choice.contains("Y")||choice.contains("y"));                    

        System.out.println("Program terminated.");

        scanner.close();

        System.exit(0);

扫描仪类不正确输入

2 个答案: