Question

在Windows中，SendMessage将阻止调用线程，直到处理完msg。但是在Android中，似乎sendMessage返回只是将msg放入消息队列中。那么，有没有办法与Android中的Windows SendMessage做同样的事情？

Answer 1

没有！
但是，您可以提供两个PendigIntents，这些内容将在发送和传递消息时被触发当调用sendTextMessage传递这些PendigIntents时：

//Get SmsManager
smsManager = SmsManager.getDefault();

//Define PendingIntent for sent
PendingIntent piSent = PendingIntent.getBroadcast(context, 0,new Intent("com.yourApp.SMS_SENT"),
                PendingIntent.FLAG_UPDATE_CURRENT);

//Define PendingIntent for delivered
PendingIntent piDelivered = PendingIntent.getBroadcast(context, 0,new Intent("com.yourApp.SMS_DELIVERED"),
                PendingIntent.FLAG_UPDATE_CURRENT);

//Send the message
smsManager.sendTextMessage(phoneNumber, null, messageText, piSent, piDelivered);

定义两个BroadcastReceivers来处理PendingIntents：

BroadcastReceiver for Sent：

public class SentReceiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {

        switch (getResultCode()){
            case Activity.RESULT_OK:
                Toast toast = Toast.makeText(context, "Message Sent",Toast.LENGTH_SHORT);
                toast.show();
            break;
            case SmsManager.RESULT_ERROR_GENERIC_FAILURE:
                toast = Toast.makeText(context, "Generic failure",Toast.LENGTH_LONG);
                toast.show();
            break;
            case SmsManager.RESULT_ERROR_NO_SERVICE:
                toast = Toast.makeText(context, "No service",Toast.LENGTH_LONG);
                toast.show();
            break;
            case SmsManager.RESULT_ERROR_NULL_PDU:
                toast = Toast.makeText(context, "Null PDU", Toast.LENGTH_LONG);
                toast.show();
            break;
            case SmsManager.RESULT_ERROR_RADIO_OFF:
                toast = Toast.makeText(context, "Radio off", Toast.LENGTH_LONG);
                toast.show();
            break;
        }
    }
}

BroadcastReceiver for Delivered：

public class DeliveredReceiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {

        switch (getResultCode()){

            case Activity.RESULT_OK:
                Toast toast = Toast.makeText(context, "Message Delivered", Toast.LENGTH_SHORT);
                toast.show();
            break;
            case Activity.RESULT_CANCELED:
                toast = Toast.makeText(context, "Message not Delivered", Toast.LENGTH_LONG);
                toast.show();
            break;
        }
    }
}

现在只需要注册这些接收器即可使用它们。

Answer 2

我可能没有清楚地提到我的问题。我真正想要的是让Handler.sendMessage（msg）阻塞调用线程，直到Handgate.handleMessage（）处理了msg。在搜索并阅读有关wait（）和notify（）的内容后，我只需使用信号量即可实现... 但无论如何，谢谢大家，圣诞快乐。： - ）

有没有一种方法sendMessage（msg）返回，直到msg被处理？

2 个答案: