Question

目前，我使用以下内容执行本机进程：

java.lang.Process process = Runtime.getRuntime().exec(command); 
int returnCode = process.waitFor();

假设没有等待程序返回，我希望在经过一定时间后终止。我该怎么做？

Answer 1

所有其他响应都是正确的，但使用FutureTask可以使其更加健壮和高效。

例如，

private static final ExecutorService THREAD_POOL 
    = Executors.newCachedThreadPool();

private static <T> T timedCall(Callable<T> c, long timeout, TimeUnit timeUnit)
    throws InterruptedException, ExecutionException, TimeoutException
{
    FutureTask<T> task = new FutureTask<T>(c);
    THREAD_POOL.execute(task);
    return task.get(timeout, timeUnit);
}

try {
    int returnCode = timedCall(new Callable<Integer>() {
        public Integer call() throws Exception {
            java.lang.Process process = Runtime.getRuntime().exec(command); 
            return process.waitFor();
        }
    }, timeout, TimeUnit.SECONDS);
} catch (TimeoutException e) {
    // Handle timeout here
}

如果重复执行此操作，则线程池更有效，因为它会缓存线程。

Answer 2

如果您使用的是Java 8，则可以使用新的waitFor with timeout：

Process p = ...
if(!p.waitFor(1, TimeUnit.MINUTE)) {
    //timeout - kill the process. 
    p.destroy(); // consider using destroyForcibly instead
}

Answer 3

这就是Plexus CommandlineUtils的用法：

Process p;

p = cl.execute();

...

if ( timeoutInSeconds <= 0 )
{
    returnValue = p.waitFor();
}
else
{
    long now = System.currentTimeMillis();
    long timeoutInMillis = 1000L * timeoutInSeconds;
    long finish = now + timeoutInMillis;
    while ( isAlive( p ) && ( System.currentTimeMillis() < finish ) )
    {
        Thread.sleep( 10 );
    }
    if ( isAlive( p ) )
    {
        throw new InterruptedException( "Process timeout out after " + timeoutInSeconds + " seconds" );
    }
    returnValue = p.exitValue();
}

public static boolean isAlive( Process p ) {
    try
    {
        p.exitValue();
        return false;
    } catch (IllegalThreadStateException e) {
        return true;
    }
}

Answer 4

Groovy方式

怎么样？

public void yourMethod() {
    ...
    Process process = new ProcessBuilder(...).start(); 
    //wait 5 secs or kill the process
    waitForOrKill(process, TimeUnit.SECONDS.toMillis(5));
    ...
}

public static void waitForOrKill(Process self, long numberOfMillis) {
    ProcessRunner runnable = new ProcessRunner(self);
    Thread thread = new Thread(runnable);
    thread.start();
    runnable.waitForOrKill(numberOfMillis);
}

protected static class ProcessRunner implements Runnable {
    Process process;
    private boolean finished;

    public ProcessRunner(Process process) {
        this.process = process;
    }

    public void run() {
        try {
            process.waitFor();
        } catch (InterruptedException e) {
            // Ignore
        }
        synchronized (this) {
            notifyAll();
            finished = true;
        }
    }

    public synchronized void waitForOrKill(long millis) {
        if (!finished) {
            try {
                wait(millis);
            } catch (InterruptedException e) {
                // Ignore
            }
            if (!finished) {
                process.destroy();
            }
        }
    }
}

Answer 5

根据我的要求修改了一下。超时是10秒。如果没有退出，过程将在10秒后被破坏。

public static void main(String arg[]) {

    try {
        Process p = Runtime.getRuntime().exec("\"C:/Program Files/VanDyke Software/SecureCRT/SecureCRT.exe\"");
        long now = System.currentTimeMillis(); 
        long timeoutInMillis = 1000L * 10; 
        long finish = now + timeoutInMillis; 
        while ( isAlive( p ) ) { 
            Thread.sleep( 10 ); 
            if ( System.currentTimeMillis() > finish ) {
                p.destroy();
            }
        }
    } catch (Exception err) {
        err.printStackTrace();
    }
}

public static boolean isAlive( Process p ) {  
    try {  
        p.exitValue();  
        return false;  
    } catch (IllegalThreadStateException e) {  
        return true;  
    }  
}

Answer 6

你需要一个2.线程来中断调用.waitFor（）的线程; 需要一些非平凡的同步才能使其健壮，但基本要点是：

TimeoutThread：

 Thread.sleep(timeout);
 processThread.interrupt();

ProcessThread：

  try {
      proc.waitFor(); 
    } catch (InterruptedException e) {
       proc.destroy();
    }

java native进程超时

6 个答案: