Question

我收到了这个错误信息Leaked Intent Receiver that was originally registered here，但即使出现了这种错误，应用程序仍然正常运行，导致该错误的原因是什么？这会影响应用吗？

这就是我想要做的事情：

private void sendSMS(String phoneNumber, String message) {
    String SENT = "SMS_SENT";
    String DELIVERED = "SMS_DELIVERED";

    ArrayList<PendingIntent> sentPendingIntents = new ArrayList<PendingIntent>();
    ArrayList<PendingIntent> deliveredPendingIntents = new ArrayList<PendingIntent>();

    PendingIntent sentPI = PendingIntent.getBroadcast(this, 0, new Intent(
            SENT), 0);

    PendingIntent deliveredPI = PendingIntent.getBroadcast(this, 0,
            new Intent(DELIVERED), 0);
    try {
        SmsManager sms = SmsManager.getDefault();
        ArrayList<String> mSMSMessage = sms.divideMessage(message);
        for (int i = 0; i < mSMSMessage.size(); i++) {
            sentPendingIntents.add(i, sentPI);
            deliveredPendingIntents.add(i, deliveredPI);
        }
        sms.sendMultipartTextMessage(phoneNumber, null, mSMSMessage,
                sentPendingIntents, deliveredPendingIntents);

    } catch (Exception e) {

        e.printStackTrace();
        Toast.makeText(this, "SMS sending failed...", Toast.LENGTH_SHORT)
                .show();
    }

    // ---when the SMS has been sent---
    registerReceiver(new BroadcastReceiver() {
        @Override
        public void onReceive(Context arg0, Intent arg1) {
            switch (getResultCode()) {
            case Activity.RESULT_OK:
                Toast.makeText(getBaseContext(), "SMS sent",
                        Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_GENERIC_FAILURE:
                Toast.makeText(getBaseContext(), "Generic failure",
                        Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_NO_SERVICE:
                Toast.makeText(getBaseContext(), "No service",
                        Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_NULL_PDU:
                Toast.makeText(getBaseContext(), "Null PDU",
                        Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_RADIO_OFF:
                Toast.makeText(getBaseContext(), "Radio off",
                        Toast.LENGTH_SHORT).show();
                break;
            }
        }
    }, new IntentFilter(SENT));

    // ---when the SMS has been delivered---
    registerReceiver(new BroadcastReceiver() {
        @Override
        public void onReceive(Context arg0, Intent arg1) {
            switch (getResultCode()) {
            case Activity.RESULT_OK:
                Toast.makeText(getBaseContext(), "SMS delivered",
                        Toast.LENGTH_SHORT).show();
                break;
            case Activity.RESULT_CANCELED:
                Toast.makeText(getBaseContext(), "SMS not delivered",
                        Toast.LENGTH_SHORT).show();
                break;
            }
        }
    }, new IntentFilter(DELIVERED));
}

提前tnx ..

Answer 1

如果您已在onCreate注册了接收者，请在其onDestroy方法中取消注册。
如果您已在onResume注册了接收者，请在其onPause方法中取消注册。
如果您已在onStart注册了接收者，请在其onStop方法中取消注册。

Answer 2

问题可能是因为您正在调用Activity的{{1}}已退出，但这是您将接收者注册为匿名内部类的上下文。您需要将这些接收器拉出，或保留对它们的引用，并在sendSMS中取消注册它们。

Answer 3

由于接收者是在活动内而不是在清单上注册的，因此您需要在onPause或onStop上取消注册

 public void onStop() {
        super.onStop();
        if (myReceiver != null) {
            unregisterReceiver(myReceiver);
        }

我忘了提到你应该在内部类中创建BroadcastReceiver，比如

  private class myReceiver extends BroadcastReceiver{

        @Override
        public void onReceive(Context context, Intent intent) {
            //Add your code here
        }
    }

希望它有所帮助。

我如何在Android中取消注册接收器？

3 个答案: