Question

我有一个发送短信（通过AsyncTask）的服务。我需要借助BraodcaseReceiver捕获发送/接收状态。由于某些原因，我无法从Receiver获取正确的resultCode。如果我在Service中实现onReceive，则resultCode始终为null。

不确定我做错了什么，所以问题是将消息从Broadcase Receiver类发送回后台服务的正确方法是什么？

背景：

该应用程序由一项活动组成，该活动只运行后台服务。在该服务中，我发送短信。

我不太擅长Android设计模式，但是从网上看到的情况来看，要获得短信状态的结果，我需要创建一个单独的类作为BraodcaseReceiver。这就是为什么我创建一个。该BroadcastReceiver应该获得已发送消息的状态，而我想要的是能够将该值传递给服务（此后从服务传递到活动）

所以我正在做的是：Aactivty-> Service-> SendSMS，然后在任务完成后，我想接收已发送短信的状态。

注意：我刚刚意识到我没有使用AsyncTask发送短信。我只想将服务用作发送sms消息的AsyncTasks的管理器，并且我不希望他们发送消息来阻止该服务。这是一个好的设计吗？

服务：

protected boolean sendSMS(String number) {

    String SENT = "SMS_SENT";
    String DELIVERED = "SMS_DELIVERED";
    PendingIntent sentPI = PendingIntent.getBroadcast(this, 0, new Intent(SENT), 0);
    PendingIntent deliveredPI = PendingIntent.getBroadcast(this, 0, new Intent(DELIVERED), 0);

    registerReceiver(sendBroadcastReceiver, new IntentFilter(SENT));
    registerReceiver(deliveryBroadcastReceiver, new IntentFilter(DELIVERED));

    String destinationAddress = number;
    String smsMessage  = String.format("This is test");
    String scAddress = null;
    SmsManager smsManager = SmsManager.getDefault();
    smsManager.sendTextMessage(destinationAddress, scAddress, smsMessage, sentPI, deliveredPI);
    return true;
}

BroadcastReceiver sendBroadcastReceiver = new BroadcastReceiver() {
    @Override
    public void onReceive(Context context, Intent intent) {
        Integer resultCode =  intent.getExtras().getInt("msg");
        //resultCode is always null here
        Log.d("Debug", "sendBroadcastReceiver code: "+ resultCode);  
    }
};

广播接收者：

public class SentReceiver extends BroadcastReceiver {
    @Override
    public void onReceive(Context context, Intent arg1) {
        Integer resultcode = getResultCode();
        //**for some reason when I define onReceive in Service class**
        //**I don't get any debug message as if this code doesn't execute**
        Log.d("Debug", "Code: "+ resultcode);

        Intent intent = new Intent("SMS_SENT");
        intent.putExtra("msg", resultcode);
        context.sendBroadcast(intent);
    }
}

从BroadcastReceiver向服务发送消息

0 个答案: