Unique Serial Number/Identification of Mobile

H friends, I am developing an official app for my organization which has confidential data to share. I want to allow only unique mobiles to use that android apk file through server-side coding. Is there any parameter in windev mobile 24/25 to detect any unique serial number (Like HDD serial of PC) of mobile and allow that serial number on a server-side database? In case the serial number is not allowed in the database, the android mobile will not be allowed to access confidential data?

 

thanks

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. GetGUID function generate distict strings during each execution so can not be linked.

 

I used following but still I can't get IMEI although get other information about phone like model/manufacturer/kernal etc..


import android.app.Activity;
import android.content.Context;
import android.content.Intent;
import android.content.SharedPreferences;
import android.content.SharedPreferences.Editor;
import android.os.*;
import android.os.Bundle;
import android.telephony.TelephonyManager;
import android.util.Log;
import android.widget.Toast;
import android.provider.Settings.Secure;

public static String getDeviceName() {

String manufacturer = android.os.Build.MANUFACTURER;
String model = android.os.Build.MODEL;
String brand = android.os.Build.BRAND;
String product = android.os.Build.PRODUCT;
String board = android.os.Build.BOARD;
String bootloader = android.os.Build.BOOTLOADER;
String cpu_abi = android.os.Build.CPU_ABI;
String cpu_abi2 = android.os.Build.CPU_ABI2;
String device = android.os.Build.DEVICE;
String display = android.os.Build.DISPLAY;
String fingerprint = android.os.Build.FINGERPRINT;
String hardware = android.os.Build.HARDWARE;
String host = android.os.Build.HOST;
String id = android.os.Build.ID;
String radio = android.os.Build.RADIO;
String serial = android.os.Build.SERIAL;
String tags = android.os.Build.TAGS;
String type = android.os.Build.TYPE;
String user = android.os.Build.USER;

if (model.startsWith(manufacturer)) {
return (model);
} else {
String MarcaModeloCelular = "manufacturer: " + (manufacturer) + "; model: " + (model) + "; brand: " + (brand) + "; product: " + (product) + "; board:" + (board) + "; bootloader: " + (bootloader) + "; cpu_abi: " + (cpu_abi) + "; cpu_abi2: " + (cpu_abi2) + "; device: " + (device) + "; display: " + (display) + "; fingerprint: " + (fingerprint) + "; hardware: " + (hardware) + "; host: " + (host) + "; id: " + (id) + "; radio: " + (radio) + "; serial: " + (serial) + "; tags: " + (tags) + "; type: " + (type) + "; user: " + (user);

MarcaModeloCelular = MarcaModeloCelular.toUpperCase();

return MarcaModeloCelular;
}
}


Any suggestion to get IMEI?

 

USed :

''''''''''''''''============
import android.app.Activity;
import android.content.Context;
import android.content.Intent;
import android.content.SharedPreferences;
import android.content.SharedPreferences.Editor;
import android.os.Bundle;
import android.telephony.TelephonyManager;
import android.util.Log;
import android.widget.Toast;
import android.provider.Settings.Secure;

//Android Manisfest
//<uses-permission android:name="android.permission.READ_PHONE_STATE"/>
//Requires READ_PHONE_STATE uses-permission

public static String RetornoNumCelular1()
{

TelephonyManager tm;
tm= (TelephonyManager)getApplicationContext().getSystemService(Context.TELEPHONY_SERVICE);
String RetornoNumCelular= tm.getDeviceId();
return RetornoNumCelular;

}


I am calling RetornoNumCelular1() function as

info(RetornoNumCelular1())
but showing error:
"System message =getDeviceId: The user 10653 does not meet the requirements to access device identifiers."

 

No issue. We can try. I will post the solution as soon as I find it. Thanks

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Got a solution at:

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Using MediaDrm API for Unique identification of a Mobile:

I used the following Code (Javascript) and called as getUniqueID () in the form as and when I need to get uniqueID which is looking promising as Google API banned IMEI fetch in a conventional way.

import android.media.MediaDrmException;
import java.lang.ref.WeakReference;
import android.media.MediaDrm;
import java.security.MessageDigest;
import java.util. *;
import java.util.UUID;

public static String getUniqueID () {
UUID wideVineUuid = new UUID (-0x121074568629b532L, -0x5c37d8232ae2de13L);
try {
MediaDrm wvDrm = new MediaDrm (wideVineUuid);
byte [] wideVineId = wvDrm.getPropertyByteArray (MediaDrm.PROPERTY_DEVICE_UNIQUE_ID);
return Arrays.toString (wideVineId);
// String mn = Arrays.toString (wideVineId);
// return mn.replaceAll (",", "");
// return Base64.encodeToString (wideVineId, Base64.Default)
// var encodedString: String = Base64.encodeToString (id, Base64.DEFAULT)
} catch (Exception e) {
// Inspect exception
return null;
}
// Close resources with close () or release () depending on platform API
// Use ARM on Android P platform or higher, where MediaDrm has the close () method
}

 

UUID method shown above is also not reliable as every update of apk causing a change of UUID.