Reputation: 1377
Suppose I have a code where it asks the user to give 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 its TimeOutException
). How do I do that?
Upvotes: 10
Views: 17196
Reputation: 5188
I use joda-time for this kind of stuff:
maven:
<!-- Joda Time -->
<dependency>
<groupId>joda-time</groupId>
<artifactId>joda-time</artifactId>
<version>1.6.2</version>
</dependency>
When prompting to input, set a LocalDateTime variable:
LocalDateTime timeOut = new LocalDateTime().plusSeconds(15);
And loop until user either inputs or the timeout is reached:
if (timeOut.isBefore(new LocalDateTime())) {
//throw your exception if this case happens
}
Upvotes: 1
Reputation: 213
How about something as simple as this:
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 throw an exception, only inform the user about being too late with his answer. (related to another question that was referred to this post).
Upvotes: -4
Reputation: 12054
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..." );
}
}
Upvotes: 5