eclipse 输入的时间限制
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Time limit for an input
提问by hari
Suppose I have a code where it asks the user to give some input, something like this:
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
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?
但我想给用户 60 秒的时间限制，然后抛出异常（在这种情况下，我认为是TimeOutException）。我怎么做？
采纳答案by Nirmal- thInk beYond
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..." );
}
}
回答by chzbrgla
I use joda-time for this kind of stuff:
我将 joda-time 用于此类内容：
maven:
行家：
<!-- 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:
当提示输入时，设置一个 LocalDateTime 变量：
LocalDateTime timeOut = new LocalDateTime().plusSeconds(15);
And loop until user either inputs or the timeout is reached:
并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
}
Before getting a down-vote: this is just a quickie :p
在获得否决票之前：这只是一个快速：p
cheers
干杯
回答by Felix
How about something as simple as this:
像这样简单的事情怎么样：
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).
这不会抛出异常，只会通知用户他的答案为时已晚。（与此帖子中提到的另一个问题有关）。
