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Java扫描程序验证返回第二个输入

[英]Java Scanner Validation returning the second input

I have a function here where it verifies the user input if its a number or within bounds. 我这里有一个函数,它可以在数字或范围内验证用户输入。

public static int getNumberInput(){
    Scanner input = new Scanner(System.in);
    while(!Inputs.isANumber(input)){
        System.out.println("Negative Numbers and Letters are not allowed");
        input.reset();
    }
    return input.nextInt();
}


public static int getNumberInput(int bound){
    Scanner input = new Scanner(System.in);
    int val =   getNumberInput();
    if(val > bound){
        System.out.println("Maximum Input is only up to: "+ bound+" Please Try Again: ");
        input.reset();
        getNumberInput(bound);
    }
    return val;
}

Each time I call getNumberInput(int bound) method with this function 每次我使用此函数调用getNumberInput(int bound)方法

public void askForDifficulty(){
    System.out.print("Difficulty For This Question:\n1)Easy\n2)Medium\n3)Hard\nChoice: ");
    int choice = Inputs.getNumberInput(diff.length);
    System.out.println(choice);
}

if I inserted an out of bound number lets say the only maximum number is 5. the getNumberInput(int bound) will call itself again. 如果我插入一个超出范围的数字,可以说唯一的最大数字是5。getNumberInput(int bound)将再次调用自身。 and when I insert the correct or within bound value it will only return to me the first value/previous value I inserted 当我插入正确或在界限值内时,它只会向我返回我插入的第一个值/上一个值

The if in the getNumberInput(int bound) should be a while . getNumberInput(int bound)if应该是while EDIT You should also combine the two methods: 编辑您还应该结合两种方法:

public static int getNumberInput(int bound){
    Scanner input = new Scanner(System.in);
    for (;;) {
        if (!Inputs.isANumber(input)) {
            System.out.println("Negative Numbers and Letters are not allowed");
            input.reset();
            continue;
        }
        int val = getNumberInput();
        if (val <= bound) {
            break;
        }
        System.out.println("Maximum Input is only up to: "+ bound+" Please Try Again: ");
        input.reset();
    }
    return val;
}

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