Java本機進程超時

Question

目前我使用下面的執行本機進程：

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

假設，而不是等待程序返回我想終止，如果在一定的時間量已過去。我該怎麼做呢？

Answer 1

這是叢CommandlineUtils是怎麼做的：

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 2

你需要一個2線程中斷調用.waitFor（）線程;將需要 一些不平凡的同步，使之強勁，但基本是：

TimeoutThread：

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

ProcessThread：

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

Answer 3

所有其他的反應是正確的，但它可以製成使用FutureTask更健壯，更高效。

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 
} 

如果這樣反反覆覆，線程池是更有效，因爲它緩存的線程。

Answer 4

怎麼樣的Groovy方式

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

根據我的要求只是修改了一下。超時在這裏是10秒。如果沒有退出，過程將在10秒後被破壞。

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

如果您使用的是Java 8，你可以簡單地使用新waitFor with timeout：

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