Java Scanner验证返回第二个输入 [英] Java Scanner Validation returning the second input

问题描述

我在这里有一个函数，它验证用户输入是否为数字或在边界内。

I have a function here where it verifies the user input if its a number or within bounds.

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;
}

每次用此函数调用getNumberInput（int bound）方法

Each time I call getNumberInput(int bound) method with this function

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);
}

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。当我插入正确的或在限定值内时，它只会返回我插入的第一个值/上一个值

if I inserted an out of bound number lets say the only maximum number is 5. the getNumberInput(int bound) will call itself again. and when I insert the correct or within bound value it will only return to me the first value/previous value I inserted

推荐答案

如果 getNumberInput（int bound）中的应为，而。 编辑您还应该合并这两种方法：

The if in the getNumberInput(int bound) should be a while. EDIT You should also combine the two methods:

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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