15

我来自 iphone 开发，如果不要求用户确认发送，您将无法在后台发送 SMS。可以在 android 的后台发送短信，这样就不需要用户干预了吗？

4

4 回答 4

28

发送带有 SMS-Delivery 通知的 SMS 作为 toast。

方法调用如下。

sendSMS("98********","This is test message");

方法签名如下。

/*
 * BroadcastReceiver mBrSend; BroadcastReceiver mBrReceive;
 */
private void sendSMS(String phoneNumber, String message) {
    ArrayList<PendingIntent> sentPendingIntents = new ArrayList<PendingIntent>();
    ArrayList<PendingIntent> deliveredPendingIntents = new ArrayList<PendingIntent>();
    PendingIntent sentPI = PendingIntent.getBroadcast(mContext, 0,
            new Intent(mContext, SmsSentReceiver.class), 0);
    PendingIntent deliveredPI = PendingIntent.getBroadcast(mContext, 0,
            new Intent(mContext, SmsDeliveredReceiver.class), 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(getBaseContext(), "SMS sending failed...",Toast.LENGTH_SHORT).show();
    }

}

现在还有两个类 SmsDeliveredReceiver、SmsSentReceiver 如下。

public class SmsDeliveredReceiver extends BroadcastReceiver {
    @Override
    public void onReceive(Context context, Intent arg1) {
        switch (getResultCode()) {
            case Activity.RESULT_OK:
                Toast.makeText(context, "SMS delivered", Toast.LENGTH_SHORT).show();
                break;
            case Activity.RESULT_CANCELED:
                Toast.makeText(context, "SMS not delivered", Toast.LENGTH_SHORT).show();
                break;
        }
    }
}

现在是 SMSSentReceiver。

public class SmsSentReceiver extends BroadcastReceiver {
    @Override
    public void onReceive(Context context, Intent arg1) {
        switch (getResultCode()) {
            case Activity.RESULT_OK:
                Toast.makeText(context, "SMS Sent", Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_GENERIC_FAILURE:
                Toast.makeText(context, "SMS generic failure", Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_NO_SERVICE:
                Toast.makeText(context, "SMS no service", Toast.LENGTH_SHORT)
                .show();
                break;
            case SmsManager.RESULT_ERROR_NULL_PDU:
                Toast.makeText(context, "SMS null PDU", Toast.LENGTH_SHORT).show();
                break;
            case SmsManager.RESULT_ERROR_RADIO_OFF:
                Toast.makeText(context, "SMS radio off", Toast.LENGTH_SHORT).show();
                break;
        }
    }
}

现在权限打开您的 AndroidManifest.xml 并添加以下行

<uses-permission android:name="android.permission.SEND_SMS"/>

它完成了.......

于 2012-11-21T10:17:54.413 回答
19

是的，您可以使用：

SmsManager sm = SmsManager.getDefault(); 
sm.sendTextMessage(number, null, message, null, null);
于 2011-06-15T17:10:34.473 回答
2

请参阅http://thinkandroid.wordpress.com/2010/01/08/sending-sms-from-application/。您将需要适当的权限。

于 2011-06-15T17:03:20.387 回答
1

最佳答案很好，但在 API 级别 23 以上，您需要务实地获得许可。否则每次都会提示权限。

 private static final int PERMISSION_REQUEST_CODE = 1;

if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.M) {

    if (checkSelfPermission(Manifest.permission.SEND_SMS)
            == PackageManager.PERMISSION_DENIED) {

        Log.d("permission", "permission denied to SEND_SMS - requesting it");
        String[] permissions = {Manifest.permission.SEND_SMS};

        requestPermissions(permissions, PERMISSION_REQUEST_CODE);

    }
}
于 2018-01-23T12:45:13.583 回答