android - 我们可以使用 WorkManager 在其他 UniqueWork 的队列中添加另一个 UniqueWork

Question

假设我已经用 WorkManager 开始了这个 UniqueWork :-

var workContinuation: WorkContinuation = WorkManager.getInstance()
                .beginUniqueWork(AppConstants.FILE_CREATOR_TAG,
                        ExistingWorkPolicy.APPEND,
                        firstOneTimeWorkRequest)

此时，我们的应用程序用户可能想要为其他任务启动另一个 UniqueWork 工作，但该任务应添加到上述工作任务的队列中。

var workContinuation: WorkContinuation = WorkManager.getInstance()
                    .beginUniqueWork(AppConstants.FILE_DISTRIBUTE_TAG,
                            ExistingWorkPolicy.APPEND,
                            fileDistrubuteTimeWorkRequest)

现在我如何告诉 WorkManager 这个 UniqueWork AppConstants.FILE_DISTRIBUTE_TAG 应该在这个 AppConstants.FILE_CREATOR_TAG 的队列中。

我不想使用 Observable 解决方案，在该解决方案中我可以跟踪此 AppConstants.FILE_CREATOR_TAG 工作人员完成并启动 AppConstants.FILE_DISTRIBUTE_TAG。

Answer 1

您应该为两个工作人员指定相同的 uniqueWorkName。引用 beginUniqueWork documentation :

The uniqueWorkName uniquely identifies this set of work.

If this method determines that new work should be enqueued and run, all records of previous work with uniqueWorkName will be pruned. If this method determines that new work should NOT be run, then the entire chain will be considered a no-op.

If any work in the chain fails or is cancelled, all of its dependent work inherits that state and will never run. This is particularly important if you are using APPEND as your ExistingWorkPolicy.

因此，你的代码应该这样写:

var workContinuation: WorkContinuation = WorkManager.getInstance()
                .beginUniqueWork(AppConstants.SOME_UNIQUE_NAME,
                        ExistingWorkPolicy.APPEND,
                        firstOneTimeWorkRequest)

// ...

var workContinuation: WorkContinuation = WorkManager.getInstance()
                .beginUniqueWork(AppConstants.SOME_UNIQUE_NAME,
                        ExistingWorkPolicy.APPEND,
                        fileDistrubuteTimeWorkRequest)