8

当我等到我的 spark apache 工作完成但没有成功时，我试图避免“while(true)”解决方案。

我有 spark 应用程序，它假设处理一些数据并将结果放入数据库，我确实从我的 spring 服务中调用它，并希望等到工作完成。

例子：

带有方法的启动器：

@Override
public void run(UUID docId, String query) throws Exception {
    launcher.addAppArgs(docId.toString(), query);

    SparkAppHandle sparkAppHandle = launcher.startApplication();

    sparkAppHandle.addListener(new SparkAppHandle.Listener() {
        @Override
        public void stateChanged(SparkAppHandle handle) {
            System.out.println(handle.getState() + " new  state");
        }

        @Override
        public void infoChanged(SparkAppHandle handle) {
            System.out.println(handle.getState() + " new  state");
        }
    });

    System.out.println(sparkAppHandle.getState().toString());
}

如何正确等待处理程序的状态为“完成”。

4

2 回答 2

6

我也在使用 Spring 应用程序中的 SparkLauncher。这是我采用的方法的摘要（通过遵循 JavaDoc 中的示例）。

用于启动作业的@Service 也实现了 SparkHandle.Listener 并通过 .startApplication 传递对自身的引用，例如

...
...
@Service
public class JobLauncher implements SparkAppHandle.Listener {
...
...
...
private SparkAppHandle launchJob(String mainClass, String[] args) throws Exception {

    String appResource = getAppResourceName();

    SparkAppHandle handle = new SparkLauncher()
        .setAppResource(appResource).addAppArgs(args)
        .setMainClass(mainClass)
        .setMaster(sparkMaster)
        .setDeployMode(sparkDeployMode)
        .setSparkHome(sparkHome)
        .setConf(SparkLauncher.DRIVER_MEMORY, "2g")
        .startApplication(this);

    LOG.info("Launched [" + mainClass + "] from [" + appResource + "] State [" + handle.getState() + "]");

    return handle;
}

/**
* Callback method for changes to the Spark Job
*/
@Override
public void infoChanged(SparkAppHandle handle) {

    LOG.info("Spark App Id [" + handle.getAppId() + "] Info Changed.  State [" + handle.getState() + "]");

}

/**
* Callback method for changes to the Spark Job's state
*/
@Override
public void stateChanged(SparkAppHandle handle) {

    LOG.info("Spark App Id [" + handle.getAppId() + "] State Changed. State [" + handle.getState() + "]");

}

使用这种方法，可以在状态变为“FAILED”、“FINISHED”或“KILLED”时采取行动。

我希望这些信息对您有所帮助。

于 2016-04-01T14:49:05.827 回答
3

我使用 CountDownLatch 实现，它按预期工作。

    ...
final CountDownLatch countDownLatch = new CountDownLatch(1);
SparkAppListener sparkAppListener = new SparkAppListener(countDownLatch);
SparkAppHandle appHandle = sparkLauncher.startApplication(sparkAppListener);
Thread sparkAppListenerThread = new Thread(sparkAppListener);
sparkAppListenerThread.start();
long timeout = 120;
countDownLatch.await(timeout, TimeUnit.SECONDS);    
    ...

private static class SparkAppListener implements SparkAppHandle.Listener, Runnable {
    private static final Log log = LogFactory.getLog(SparkAppListener.class);
    private final CountDownLatch countDownLatch;
    public SparkAppListener(CountDownLatch countDownLatch) {
        this.countDownLatch = countDownLatch;
    }
    @Override
    public void stateChanged(SparkAppHandle handle) {
        String sparkAppId = handle.getAppId();
        State appState = handle.getState();
        if (sparkAppId != null) {
            log.info("Spark job with app id: " + sparkAppId + ",\t State changed to: " + appState + " - "
                    + SPARK_STATE_MSG.get(appState));
        } else {
            log.info("Spark job's state changed to: " + appState + " - " + SPARK_STATE_MSG.get(appState));
        }
        if (appState != null && appState.isFinal()) {
            countDownLatch.countDown();
        }
    }
    @Override
    public void infoChanged(SparkAppHandle handle) {}
    @Override
    public void run() {}
}
于 2016-11-16T04:01:42.460 回答