Question

我想在游戏中加入战斗，我将如何制作它以便用户只有特定的时间来输入一个键 - 组合ex

String [] arr = {＆＃34; A＆＃34;，＆＃34; B＆＃34;，＆＃34; C＆＃34;，＆＃34; D＆＃34;，＆＃34; E＆＃34;，＆＃34; F＆＃34;}; 随机随机= new Random（）;

    int fight = random.nextInt(arr.length);
    if (arr[fight].equals("A")) {
        System.out.println("You encountered an enemy!, Get ready to fight!");
        try {
            Thread.sleep(2000);
        } catch (InterruptedException f) {
        }
        System.out.println("The enemy shoots a bullet!");
        System.out.print("Enter \"10110\" to dodge!: "); ---------
        pressp = i.next();-------- only 1 second to enter "10110"
        try {                             else you get shot by bullet.
            Thread.sleep(1000);
        } catch (InterruptedException f) {
        }

Answer 1

您当前的代码将在扫描仪上的下一个方法调用中被阻止，因此您无法获得您正在寻找的内容。一旦你输入一些字符串，它只会睡一秒钟。您可能需要的是：

//define a callable which will be executed asynchronously in a thread.
static class MyString implements Callable<String> {
    Scanner scanner;
    public MyString(Scanner scanner) {
        this.scanner = scanner;
    }

    @Override
    public String call() throws Exception {
        System.out.println("Enter now");
        return scanner.next();
    }
}

public static void main(String[] args) {
    Scanner userInput = new Scanner(System.in);
    ExecutorService executor = Executors.newFixedThreadPool(1);//just one thread
    MyString str = new MyString(userInput);
    Future<String> future = executor.submit(str);//it will ask user for input asynchronously and will return immediately future object which will hold string that user enters.
    try {
        String string = future.get(1, TimeUnit.SECONDS);//get user input within a second else you will get time out exception.
        System.out.println(string);
    } catch (InterruptedException | ExecutionException | TimeoutException e) {
        e.printStackTrace();//if user failed to enter within a second..
    }
}

花一定的时间进入int

1 个答案: