如何在此循环中合并'int only'输入？

Question

我已经花了大约一个小时的时间来处理这个代码，现在我试图让它在用户输入字符而不是数字时循环

    System.out.println("Enter level student last completed (0-3): "); 
    int level = in.nextInt();
    while (level > 3 || level < 0){
        System.out.println("Please enter a valid level!: ");
        level = in.nextInt();
    }

我想在while (level > 3 || level < 0)行中添加!in.hasNextInt()，使其成为

while (!in.hasNextInt() || level > 3 || level < 0)

但这并没有帮助，因为如果输入了字符，程序仍然会崩溃。

编辑：

    System.out.println("Enter level student last completed (0-3): "); 
    int level = 1; //in.nextInt();
    while (in.hasNextInt()==false || level > 3 || level < 0){
        in.next();
        System.out.println("Please enter a valid level!: ");
    }
    level = in.nextInt();

Answer 1

您可以创建一个while循环来重复请求输入，直到它得到它喜欢的

基本解决方案

    int i;
    while(scan.hasNextInt()==false){ //keep asking until it gets something it likes
        scan.next(); //<--consume bad input, important!
        System.out.println("Only integers are valid");
    }
    i=scan.nextInt();
    
    System.out.println(i);

高级解决方案

你可以把它打包成一种方法，当我们想加入更多的逻辑时，这种方法会让生活变得更容易

public static int getSafeInteger(){
    Scanner scan=new Scanner(System.in); //if using scanner over and over consider passing the scanner as an argument
    while(scan.hasNextInt()==false){
        scan.next();
        System.out.println("Only integers are valid");
    }
    return scan.nextInt();
}

然后我们可以在您现有的环路中使用该方法

    System.out.println("Enter level");
    
    Scanner scan=new Scanner(System.in);
    
    int level=getSafeInteger();
    while (level > 3 || level < 0){
        System.out.println("Please enter a valid level!: ");
        level = getSafeInteger();
    }

Answer 2

使用nextLine()，并尝试使用Integer.parseInt(String)将其解析为整数。

Answer 3

您可以这样做：

int level;
while (true) {
    if (in.hasNextInt()) {
        // now we know at least the input is a number
        if (level > 3 || level < 0) {
            // aww, it's an invalid number :(
            in.nextLine(); // clear bad input
            System.out.println("Please enter a valid level (0-3)!: ");
        } else {
            // hooray! store the input now
            level = in.nextInt();
            break; // out of the infinite loop
        }
    } else {
        // input is not a number
        in.nextLine(); // clear bad input
        System.out.println("Please enter a number!: ");
    }
}

Answer 4

试试这个：

Scanner sc = new Scanner(System.in);
int level;
System.out.printf("Enter level student last completed (0-3): "); 
while(true) {
    try {
        level = Integer.parseInt(sc.next());
        if (level < 4 && level > -1) {
            break;
        }
    } catch (Exception ex) { }
    System.out.printf("Please enter a valid level!: ");
}

Answer 5

您可以设置级别的默认值大于3并执行以下

            do
            {
                System.out.println("Enter level student last completed (0-3): ");
                try{
                 level = in.nextInt();
                }catch(InputMismatchException e){
                    System.out.println("Not a valid Number");
                    in.nextLine();
                }
            }while (level > 3 || level < 0);