5

我已经创建了IntentService类并执行asyncTask但onPreExecute()在此代码行调用时出现异常pDialog.show();

AsyncHandlerService 类 ---

public class AsyncHandlerService extends IntentService{
ProgressDialog pDialog;
HttpPost post;
HttpResponse response;
Context ctx;

public AsyncHandlerService() {
    super("AsyncHandlerService");
    ctx = this;
}

@Override
protected void onHandleIntent(Intent intent) {
    new LoadDeviceInfo().execute();   
}


class LoadDeviceInfo extends AsyncTask<String, String, String> {

@Override
protected void onPreExecute() {
    super.onPreExecute();
    pDialog = new ProgressDialog(ctx);
    pDialog.setMessage("Updating device info...");
    pDialog.setIndeterminate(false);
    pDialog.setCancelable(false);
    pDialog.show(); //Exception here..
}

protected String doInBackground(String... args) {
}

protected void onPostExecute(String file_url) {
    pDialog.dismiss();
}

更新：

我在广播接收器中调用了在 android manifestIntentService中定义的意图过滤器。android.intent.action.PACKAGE_REPLACED编码 - -

public class OnUpgradeBroadcastReceiver extends BroadcastReceiver {
Context activity;
@Override
    public void onReceive(final Context context, final Intent intent) {
         activity = context;
         Intent msgIntent = new Intent(activity, AsyncHandlerService.class);
            activity.startService(msgIntent);
    }
}

错误日志：

com.testapp.main fatal error : Unable to add window -- 
token null is not for an application
android.view.WindowManager$BadTokenException: Unable to add window -- 
token null is not for an application
at android.view.ViewRootImpl.setView(ViewRootImpl.java:588)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:326)
at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:224)
at android.view.WindowManagerImpl$CompatModeWrapper.
addView(WindowManagerImpl.java:149)
at android.app.Dialog.show(Dialog.java:277)
at com.testapp.main.AsyncHandlerService$LoadDeviceInfo.
onPreExecute(AsyncHandlerService.java:62)
at android.os.AsyncTask.executeOnExecutor(AsyncTask.java:586)
at android.os.AsyncTask.execute(AsyncTask.java:534)
4

5 回答 5

13

首先，IntentService已经使用了后台线程。您不需要另一个后台线程。做需要在后台完成的工作onHandleIntent()。

其次，aService不能显示 aDialog。相反，让您的应用程序的 UI 层知道工作是通过事件总线上的消息完成的（例如，LocalBroadcastManagergreenrobot 的 EventBus，Square 的 Otto）。如果 UI 层不处理该事件，如果Notification需要，您的服务可以引发或以其他方式让用户知道已完成的工作。

于 2014-05-29T14:28:20.913 回答
0

服务不是 UI 线程。
由于您尝试显示ProgressDialog来自服务上下文的 a，因此无法完成。
试试这个解决方案：
https ://stackoverflow.com/a/4369755/1405268

于 2014-05-29T14:18:15.117 回答
0

如果出于某种原因您真的真的 很想使用 AsyncTask（例如，您已经设置了框架以使用 AsyncTask 来调用某些 web api），您总是可以使用等待/通知，例如：

public class GetCacheIntentService extends DebuggableIntentService implements ApiAsyncTask.Callback {
    private static final String ACTION_GET_CACHE = "action.GET_CACHE";

    private static final String EXTRA_INT_START = "extras.START";
    private static final String EXTRA_INT_LIMIT = "extras.LIMIT";

    private static final int API_GET_CACHE = 0;

    private final Object mApiCallLock = new Object();
    private GetCacheResponse getCacheResponse;

    public GetCacheIntentService() {
        super("GetCacheIntentService");
        setIntentRedelivery(true);
    }

    public static void startServiceActionGetCache(Context context, int start, int limit) {
        Intent intent = new Intent(context, GetCacheIntentService.class);
        intent.setAction(ACTION_GET_CACHE);
        intent.putExtra(EXTRA_INT_START, start);
        intent.putExtra(EXTRA_INT_LIMIT, limit);
        context.startService(intent);
    }

    @Override
    protected void onHandleIntent(Intent intent) {
        if (intent == null) {
            return;
        }

        String action = intent.getAction();

        if (ACTION_GET_CACHE.equals(action)) {
            int start = intent.getIntExtra(EXTRA_INT_START, 0);
            int limit = intent.getIntExtra(EXTRA_INT_LIMIT, 100);
            getCache(start, limit);
        }
    }

    private void getCache(int start, int limit) {
        GetCacheTask task = new GetCacheTask(this, API_GET_CACHE);
        task.setStart(start);
        task.setLimit(limit);
        task.execute();

        synchronized (mApiCallLock) {
            try {
                mApiCallLock.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
                Thread.currentThread().interrupt();
            }
        }

        processResponse(mGetCacheResponse);
    }

    public void processResponse(GetCacheResponse response) {
           // do something
    }

    @Override
    public void onRequestFailed(int id, ApiResponse apiResponse) {
        synchronized (mApiCallLock) {
            switch (id) {
                case API_GET_CACHE:
                    break;
            }
            mApiCallLock.notify();
        }
    }

    @Override
    public void onRequestSuccess(int id, ApiResponse response) {
        synchronized (mApiCallLock) {
            switch (id) {
                case API_GET_CACHE:
                    mGetCacheResponse = (GetCacheResponse) response;
                    break;
            }
            mApiCallLock.notify();
        }
    }
}

不过这很丑:(

于 2015-04-22T14:09:28.243 回答
0

从 Intent 服务调用 Asynctask 不是一个好习惯。如果您需要从 IntentService 旋转其他线程，请考虑使用 Executor。

ExecutorService es = Executors.newFixedThreadPool(5);
es.execute(new Runnable() {
                @Override
                public void run() {

                }
            });

es.execute(new Runnable() {
                @Override
                public void run() {
                }
            });
es.shutdown();
es.awaitTermination(1, TimeUnit.HOURS);
于 2016-01-15T03:00:32.560 回答
0

在 IntentService 子类甚至 JobIntentService 子类中使用 AsyncTask 不是一个好习惯。在 JobIntentServices 的情况下，它也会导致崩溃。

于 2019-09-17T16:40:48.303 回答