有没有办法从json格式的网站安全地传输数据到Android应用程序？

Question

I have website that has a database inside, I want to retrieve some really secure information from that database in the format of JSON to my android application. For example when I login to my website through my android application, the website transfer JSON securely to me. I want to make sure no one else could decode my JSON, for example if some hacker could login with user credentials into my website, he could not access this secure information. I have read some articles about JWT and I got really confused. I really appreciate you to lead me through the correct way. The problem with the json was something like this: when I add some code inside my php code to hide it from outside (only legitimate user can access that, I did it with session code) the result of json is going to be null when I connect with android I do not have this problem when I did in in windows, maybe it is something wrong with my android code when I want to pars json, but whenever I delete the session part it worked perfectley. Here is my android code

public class ParsingJSON {
public static String[] ids;
public static String[] names;
public static String[] passwords;
public static String[] fullnames;
public static String[] emails;
public static String[] otps;


public static final String JSON_ARRAY = "result";
public static final String KEY_ID = "id";
public static final String KEY_NAME = "username";
public static final String KEY_PASSWORD = "password";
public static final String KEY_FULLNAME= "fullname";
public static final String KEY_EMAIL= "email";
public static final String KEY_OTP = "otp";

private JSONArray users = null;

private String json;

public ParsingJSON(String json){

    this.json = json;
}

protected String[] ParsingJSON(){
    JSONObject jsonObject=null;
    try {

        jsonObject = new JSONObject(json);



        users = jsonObject.getJSONArray(JSON_ARRAY);

        Log.v("sharareh:", String.valueOf(users));

        ids = new String[users.length()];
        names = new String[users.length()];
        passwords = new String[users.length()];
        fullnames = new String[users.length()];
        emails = new String[users.length()];
        otps = new String[users.length()];

        for(int i=0;i<users.length();i++){
            Log.v("sharareh1:", "sharareh");
            JSONObject jo = users.getJSONObject(i);
            ids[i] = jo.getString(KEY_ID);
            names[i] = jo.getString(KEY_NAME);
            passwords[i] = jo.getString(KEY_PASSWORD);
            fullnames[i] = jo.getString(KEY_FULLNAME);
            emails[i] = jo.getString(KEY_EMAIL);
            otps[i] = jo.getString(KEY_OTP);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

    Log.v("OTPJSON:", otps[0]);
    return otps;
}

Here is my php code:

<?php
define('HOST', 'x');
define('USER', 'y');
define('PASS', 'z');
define('DB', 'y');
require_once 'totp.class.php';
session_start();
include_once 'include/class.user.php';
$user = new User();
$var_value = $_SESSION['varname'];
echo $var_value;
$user =  $var_value;
$con = mysqli_connect(HOST,USER,PASS,DB);
$sql = "select * from users WHERE uemail='$user' ";
$res = mysqli_query($con,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){ 
array_push($result,
array('id'=>$row[0],
'username'=>$row[1],
'password'=>$row[2],
'fullname'=>$row[3],
'email'=>$row[4],
'otp'=>$row[5]
));
}
echo json_encode(array("result"=>$result));
mysqli_close($con);?>