Question

我是Android新手，在我的应用中，我想在不打开现有MessageApp的情况下向手机发送短信。为实现这一目标，我编写了下面的代码，但我手机中没有收到任何消息。

有什么问题？

码

private void sendSms(String phoneNo, String msg) {
    try {
        SmsManager smsManager = SmsManager.getDefault();
        smsManager.sendTextMessage(phoneNo, null, msg, null, null);
        Toast.makeText(getApplicationContext(), "Message Sent",
                Toast.LENGTH_LONG).show();
    } catch (Exception ex) {
        Toast.makeText(getApplicationContext(), ex.getMessage().toString(),
                Toast.LENGTH_LONG).show();
        ex.printStackTrace();
    }
}

另外，我将以下权限添加到我的宣言文件中。

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

Answer 1

您必须导入：import android.telephony.SmsManager;

此外，您还需要为Api＆gt; 23

添加运行时权限

if(isSMSPermissionGranted()){
    sendSms(String phoneNo, String msg);
}

使用此方法：

    public  boolean isSMSPermissionGranted() {
    if (Build.VERSION.SDK_INT >= 23) {
        if (checkSelfPermission(android.Manifest.permission.SEND_SMS)
                == PackageManager.PERMISSION_GRANTED) {
            Log.v(TAG,"Permission is granted");
            return true;
        } else {

            Log.v(TAG,"Permission is revoked");
            ActivityCompat.requestPermissions(getActivity(), new String[]{Manifest.permission.SEND_SMS}, 0);
            return false;
        }
    }
    else { //permission is automatically granted on sdk<23 upon installation
        Log.v(TAG,"Permission is granted");
        return true;
    }
}

捕获权限结果：

 @Override
public void onRequestPermissionsResult(int requestCode,
                                       String permissions[], int[] grantResults) {
    switch (requestCode) {

        case 0: {

            if (grantResults.length > 0
                    && grantResults[0] == PackageManager.PERMISSION_GRANTED) {
                Toast.makeText(getContext(), "Permission granted", Toast.LENGTH_SHORT).show();
                //send sms here call your method
                sendSms(String phoneNo, String msg);
            } else {
                Toast.makeText(getContext(), "Permission denied", Toast.LENGTH_SHORT).show();
            }
            return;
        }

        // other 'case' lines to check for other
        // permissions this app might request
    }
 }

Answer 2

sendSMS()功能将是

    //---sends an SMS message to another device---
    private void sendSMS(String phoneNumber, String message)
    {        
        PendingIntent pi = PendingIntent.getActivity(this, 0,
            new Intent(this, SMS.class), 0);                
        SmsManager sms = SmsManager.getDefault();
        sms.sendTextMessage(phoneNumber, null, message, pi, null);        
    }

希望本教程能为您提供帮助！ https://www.tutorialspoint.com/android/android_sending_sms.htm

Answer 3

使用此功能......

String mobileNumber="0123456789";
String messege="Message";

SmsManager sms = SmsManager.getDefault();
List<String> messages = sms.divideMessage(messege);

for (String msg : messages)
{
     PendingIntent sentIntent = PendingIntent.getBroadcast(getActivity(), 0,new   Intent("SMS_SENT"), 0);
     PendingIntent deliveredIntent =PendingIntent.getBroadcast(getActivity(), 0,new  Intent("SMS_DELIVERED"), 0);

     sms.sendTextMessage(personalPhone, null, msg, sentIntent, deliveredIntent);
}

Answer 4

import android.telephony.SmsManager; 不是import android.telephony.gsm.SmsManager;

这项工作在我的手机上。版本6 +可能需要权限，因此请查看并添加权限。它会对你有所帮助。

如何在不打开消息应用的情况下以编程方式发送SMS

码

4 个答案: