Firebase jobdispatcher 未在指定窗口内触发

Posted 2023-04-14

【中文标题】Firebase jobdispatcher 未在指定窗口内触发

【英文标题】：Firebase jobdispatcher not triggering within specified window

【发布时间】：2017-11-08 08:29:05

【问题描述】：

我正在实施 Firebase Jobdispatcher，触发时间在 10 到 20 秒之间。这是我安排工作的代码：

public static void scheduleCompatibleJob(Context context) 
    FirebaseJobDispatcher dispatcher = new FirebaseJobDispatcher(new GooglePlayDriver(context));
    Job myJob = dispatcher.newJobBuilder()
            .setService(DataTrackerJobService.class) // the JobService that will be called
            .setTag("api_trigger")
            .setRecurring(true)
            .setLifetime(Lifetime.FOREVER)
            .setTrigger(Trigger.executionWindow(10, 20)) // 10 to 20 seconds
            .setReplaceCurrent(true)
            .setRetryStrategy(RetryStrategy.DEFAULT_LINEAR)
            .setConstraints(Constraint.ON_ANY_NETWORK)
            .build();

    dispatcher.mustSchedule(myJob);

还有服务类：

public class DataTrackerJobService extends JobService 
@Override
public boolean onStartJob(final JobParameters job) 
    Log.e("start job", "started " + job);
    (new Handler()).postDelayed(new Runnable() 
        @Override
        public void run() 
            Log.e("handler", "run");
            jobFinished(job, true);
        
    , 2000);
    return true;


@Override
public boolean onStopJob(JobParameters job) 
    Log.e("stop job", "stopped " + job);
    return false;

Job dispatcher 正在运行，但 logcat 中的时间不正确。随着工作的每一次重新安排，时间差距不断增加，从来没有在 10 到 20 秒之间。

06-07 11:19:16.429 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f4250de4 06-07 11:19:18.432 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:20:16.436 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f16ceb31 06-07 11:20:18.438 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:21:58.519 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f4c635cd 06-07 11:22:00.520 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:23:40.602 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f15f8e29 06-07 11:23:42.605 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:25:52.642 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f48c1045 06-07 11:25:54.644 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:28:52.652 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@f3b49821 06-07 11:28:54.655 26174-26174/com.example.jobdispatcher E/handler: 运行 06-07 11:32:04.688 26174-26174/com.example.jobdispatcher E/开始工作：开始 com.firebase.jobdispatcher.JobInvocation@e7f3c1bd 06-07 11:32:06.690 26174-26174/com.example.jobdispatcher E/handler: 运行

请在 logcat 中检查时间。指导我在哪里出错，或者它应该只以这种方式工作？基本上我想在 24 小时的时间间隔内实现它，但我想知道这是否可行，然后在触发时间中指定的时间的两倍后调用一次作业。

【参考方案1】：

如果您查看 Trigger 类的来源，您会注意到您不确定您的作业是否会在给定时间运行。

/**
 * Creates a new ExecutionWindow based on the provided time interval.
 *
 * @param windowStart The earliest time (in seconds) the job should be
 *                    considered eligible to run. Calculated from when the
 *                    job was scheduled (for new jobs) or last run (for
 *                    recurring jobs).
 * @param windowEnd   The latest time (in seconds) the job should be run in
 *                    an ideal world. Calculated in the same way as
 *                    @code windowStart.
 * @throws IllegalArgumentException if the provided parameters are too
 *                                  restrictive.
 */
public static JobTrigger.ExecutionWindowTrigger executionWindow(int windowStart, int windowEnd) 
    if (windowStart < 0) 
        throw new IllegalArgumentException("Window start can't be less than 0");
     else if (windowEnd < windowStart) 
        throw new IllegalArgumentException("Window end can't be less than window start");
    

    return new JobTrigger.ExecutionWindowTrigger(windowStart, windowEnd);

您还应该注意，下一个重复性工作只有在上一个工作结束时才开始。所以当你的工作时间长且成本高时，下一次执行时间可能会出乎意料。对于耗时的工作，您应该扩展 SimpleJobService。

为了创建一个重复的任务，我使用了我的 util 方法来创建一个适当的触发器：

public class JobDispatcherUtils 
    public static JobTrigger periodicTrigger(int frequency, int tolerance) 
        return Trigger.executionWindow(frequency - tolerance, frequency);
    

用法：

class DataTrackerJobService extends SimpleJobService 
    // ...


public static void scheduleCompatibleJob(Context context) 
    FirebaseJobDispatcher dispatcher = new FirebaseJobDispatcher(new GooglePlayDriver(context));
    Job myJob = dispatcher.newJobBuilder()
            .setService(DataTrackerJobService.class) // the JobService that will be called
            .setTag("api_trigger")
            .setRecurring(true)
            .setLifetime(Lifetime.FOREVER)
            .setTrigger(JobDispatcherUtils.periodicTrigger(20, 1)) // repeated every 20 seconds with 1 second of tollerance
            .setReplaceCurrent(true)
            .setRetryStrategy(RetryStrategy.DEFAULT_LINEAR)
            .setConstraints(Constraint.ON_ANY_NETWORK)
            .build();

    dispatcher.mustSchedule(myJob);

【讨论】：

20 秒内仍未执行

但是在小米这样的中国设备中，如果应用程序从任务栏作业计划中清除将永远无法工作