Question

假设我有一个代码，它要求用户提供一些输入，如下所示：

for (condition) {
System.out.println("Please give some input");
System.in.read();
} //lets say this loop repeats 3 times and i face a problem during second iteration

但是我想给用户一个60秒的时间限制，然后抛出异常（在这种情况下，我认为它是TimeOutException）。我该怎么做？

Answer 1

import java.util.Timer;
import java.util.TimerTask;
import java.io.*;
public class test
{
    private String str = "";

    TimerTask task = new TimerTask()
    {
        public void run()
        {
            if( str.equals("") )
            {
                System.out.println( "you input nothing. exit..." );
                System.exit( 0 );
            }
        }    
    };

    public void getInput() throws Exception
    {
        Timer timer = new Timer();
        timer.schedule( task, 10*1000 );

        System.out.println( "Input a string within 10 seconds: " );
        BufferedReader in = new BufferedReader(
        new InputStreamReader( System.in ) );
        str = in.readLine();

        timer.cancel();
        System.out.println( "you have entered: "+ str ); 
    }

    public static void main( String[] args )
    {
        try
        {
            (new test()).getInput();
        }
        catch( Exception e )
        {
            System.out.println( e );
        }
        System.out.println( "main exit..." );
    }
}

Answer 2

我使用joda-time来做这种事情：

行家：

  <!--  Joda Time -->
    <dependency>
        <groupId>joda-time</groupId>
        <artifactId>joda-time</artifactId>
        <version>1.6.2</version>
    </dependency>

提示输入时，设置LocalDateTime变量：

 LocalDateTime timeOut = new LocalDateTime().plusSeconds(15);

循环直到用户输入或达到超时：

 if (timeOut.isBefore(new LocalDateTime())) {
 //throw your exception if this case happens
 }

在进行投票之前：这只是一个快速：p

欢呼声

Answer 3

如此简单的事情：

    Scanner reader = new Scanner(System.in); 
    System.out.println("Enter a number: ");
    long limit = 5000L;
    long startTime = System.currentTimeMillis();
    Long l = reader.nextLong();
    if ((startTime + limit) < System.currentTimeMillis())
        System.out.println("Sorry, your answer is too late");
    else
        System.out.println("Your answer is on time");

这不会引发异常，只会告诉用户他的回答太迟了。（与提到这篇文章的另一个问题有关）。

输入的时间限制

3 个答案: