Time limit for input

Question

Suppose I have a code where it asks the user for some input, something like this:

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

but I want to give the user a 60 second time limit and then throw an exception (in this case, I think it is TimeOutException). How to do it?

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

I am using joda-time for this kind of thing:

Maven:

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

When prompted for input, enter the LocalDateTime variable:

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

And loop until the user enters an input or timeout:

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

Before you start voting: this is just quickie: p

amuses

Answer 3

- :

    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");

This will not lead to an exception, only inform the user about being late with his answer. (related to another question that was related to this post).