java为什么是扫描仪。下一步（）不接受键盘回车键？

Question

import java.util.InputMismatchException;
import java.util.Scanner;

class ClassB {

    public static void main(String[]args) throws Exception{
        ClassB b = new ClassB();
        b.getInput();
    }

    public void getInput() throws Exception {
        String label = null;
        int input;

        Scanner scan = new Scanner(System.in);
        for (input = 1; input != 4; input++){
            switch(input){
            case 1: label = "name";
            break;
            case 2: label = "password";
            break;
            case 3: label = "Room number";
            break;
            }
            System.out.println("Enter your " + label);
            scan.next();
            try{
                if (input == 1){
                    int name = scan.nextInt();
                    //residence.changeName(name);       
                }
                else if(input == 2){
                    String password = scan.next();
                }
                int rmNumber = scan.nextInt();
            }catch (IllegalArgumentException | InputMismatchException  me ){
                String type = "A string";
                String message = (input == 1 || input == 2) ? type : "An integer";
                input = 1;
            }
        }
    }
}

当我运行此代码时，第一次扫描。当按下计算机的enter键时，try catch块中的next（）不响应，因此无法输入后续的if（）语句。文本字段中的光标只换行，但永远不会接受输入。 我正在使用NetBeanIDE

Answer 1

从您的回答来看，您仍然看到它被卡住，并且要求用户多次输入值，我不确定代码的区别是什么。这是我的代码，我用它进行了测试，但没有遇到这个问题（对我来说）

import java.util.Scanner;

public class TestInput{
    private void display(String s){
        System.out.println("Please input a valid value: " + s);
    }

    public void getInput(Residence residence){
        String label = null;
        int input;
        Scanner scan = new Scanner(System.in);

        for (input = 1; input != 4; input++){

            switch(input){
                case 1: label = "name";
                break;
                case 2: label = "password";
                break;
                case 3: label = "Room number";
               break;
            }

            System.out.println("Enter your " + label);

            try{
                if (input == 1){
                    residence.changeName(scan.next());       
                }
                else if(input == 2){
                   String password = scan.next();
                   residence.changePassword(password);
                }
                else if(input == 3){
                   int roomNumber = scan.nextInt();
                   residence.changeRoomNumber(roomNumber);
                }
            }
            catch (Exception me ){
                String type = "A string";
                String message = (input == 1 || input == 2) ? type : "An integer";
                display(message);
                input = 1;
            }
        }
   }

   public static void main(String... args){
    TestInput ti = new TestInput();
    ti.getInput(new Residence());
   }
}

class Residence{
    String password;
    String name;
    int room;

    public void changeName(String n){
        this.name = n;
    }
    public void changePassword(String pass){
        this.password = pass;
    }
    public void changeRoomNumber(int n){
        room = n;
    }
}

在try/catch if/else语句中，您没有检查条件3。 您被“卡住”的原因是您还第二次调用了nextInt（），导致它在第一次输入房间号后仍等待其他输入。将if条件替换为以下条件，然后重试：

           if (input == 1){
                residence.changeName(scan.next());       
            }
            else if(input == 2){
                String password = scan.next();
                residence.changePassword(password);
            }
            else if(input == 3){
                int roomNumber = scan.nextInt();
                residence.changeRoomNumber(roomNumber);
            }

Answer 2

http://javatutorialhq.com/java/util/scanner-class-tutorial/

nextInt method does not read the last newline character of your input, and thus that newline is consumed in the next call to nextLine. To resolve this just use next instead of nextline but if you insist of using nextLine add another scan.nextLine() after nextInt. Take a look below snippet

->；添加扫描。扫描后的下一行（）。nextInt（）