Java - 如果语句逻辑错误

Question

事件的顺序应为：

  

1）我输入两个。
  2）我被提示选择一个名字   3）我选择华盛顿/富兰克林/汉密尔顿   4）我被问到这个名字出现在哪个面额上   5）我给出答案。

但是，当我进入华盛顿参加第三部分时 - 我被告知这是一个无效的数字。我不明白为什么会这样。

public static void main(String[] args) {
    // TODO code application logic here
    System.out.println("Type 1 to enter a denomination, 2 to enter a last name");
    Scanner in = new Scanner(System.in);
    int x = in.nextInt();

    if(x==1){
        System.out.println("Choose a denomination");
        int y = in.nextInt();
        in.nextLine();
        if(y==1){
            System.out.println("Which person appears on the 1 bill?");
            String answer = in.nextLine();
            if(answer.equals("Washington")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }


        else if(y==10){
            System.out.println("Which person appears on the 10 bill?");
            String answer = in.nextLine();
            if(answer.equals("Hamilton")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else if(y==100){
            System.out.println("Which person appears on the 100 bill?");
            String answer = in.nextLine();
            if(answer.equals("Franklin")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else{
            System.out.println("That is an invalid number.");
        }
    }
    else if(x==2){
       System.out.println("Choose a name");
        String y = in.nextLine();
        in.nextLine();
        if(y.equals("Washington")){
            System.out.println("Which denomination does this name appear on?");
            int answer = in.nextInt();
            if(answer==1){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }


        else if(y.equals("Hamilton")){
            System.out.println("Which denomination does this name appear on");
            int answer = in.nextInt();
            if(answer==10){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else if(y.equals("Franklin")){
            System.out.println("Which denomination does this name appear on");
            int answer = in.nextInt();
            if(answer==100){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else{
            System.out.println("That is an invalid number.");
        }
    } 
}

问题在于x == 2段。 x == 1工作正常。

Answer 1

问题不依赖于看似额外的in.nextLine()。

我会建议您将所有in.nextInt()更改为in.nextLine() ，然后将其解析，改为实际类型，例如int

示例：

int answer = Integer.parseInt(in.nextLine());

建议更改的原因：当您使用nextInt()时，有一种趋势是，下一行仍然会在那里徘徊，直到您执行额外的nextLine()清除它。 / p>

为防止出现此类问题，建议您使用nextLine()接收所有输入，然后将其解析为实际类型（int，double等。）

这也是 Microsoft 处理C#中整数输入的方式。

已编辑的工作代码：

public static void main(String[] args) {

    System.out.println("Type 1 to enter a denomination, 2 to enter a last name");
    Scanner in = new Scanner(System.in);
    int x = Integer.parseInt(in.nextLine());

    if(x==1){
        System.out.println("Choose a denomination");
        int y = Integer.parseInt(in.nextLine());
        in.nextLine();
        if(y==1){
            System.out.println("Which person appears on the 1 bill?");
            String answer = in.nextLine();
            if(answer.equals("Washington")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }


        else if(y==10){
            System.out.println("Which person appears on the 10 bill?");
            String answer = in.nextLine();
            if(answer.equals("Hamilton")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else if(y==100){
            System.out.println("Which person appears on the 100 bill?");
            String answer = in.nextLine();
            if(answer.equals("Franklin")){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else{
            System.out.println("That is an invalid number.");
        }
    }
    else if(x==2){
       System.out.println("Choose a name");
        String y = in.nextLine();
        //in.nextLine();
        if(y.equals("Washington")){
            System.out.println("Which denomination does this name appear on?");
            int answer = Integer.parseInt(in.nextLine());
            if(answer==1){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }


        else if(y.equals("Hamilton")){
            System.out.println("Which denomination does this name appear on");
            int answer = Integer.parseInt(in.nextLine());
            if(answer==10){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else if(y.equals("Franklin")){
            System.out.println("Which denomination does this name appear on");
            int answer = Integer.parseInt(in.nextLine());
            if(answer==100){
                System.out.println("That is correct");
            }
            else{
                System.out.println("That is incorrect");
            }
        }
        else{
            System.out.println("That is an invalid number.");
        }
    } 
}

Answer 2

String y = in.nextLine();
in.nextLine();
if(y.equals("Washington")){

您多次调用in.nextLine（）。删除第二行。

Answer 3

您必须在nextInt()之后跳过该行，然后接受用户输入（在y==2情况下），因为nextInt()不会消耗它使用该令牌的整行。

in.nextLine();
String y = in.nextLine();

使用您的代码，in.nextLine()中的String y = in.nextLine();正在读取您输入x值的同一行，其值y为""，因此输出"That is an invalid number"

Answer 4

当您执行String y = in.nextLine()时，下一个标记是上一次nextInt()来电中的剩余换行符，因此y最终等于\n。

使用nextLine()后打电话给nextInt()：

int x = in.nextInt();
in.nextLine();

之后，您应该只能在获得用户输入时拨打nextLine一次：

String y = in.nextLine();
if(y.equals("Washington")){