No JSON response from PHP - php

Question

I have a JSON POST request being sent to index.php as part of a login application for a mobile device. I'm using an old script so I believe the problem is deprecated syntax with the PHP, as my response JSON is coming back empty. The $user below isn't doing anything as I'm just debugging.
index.php
if (isset($_POST['tag']) && $_POST['tag'] != '') {
// get tag
$tag = $_POST['tag'];
// include db handler
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
// response Array
//$response = array('tag' => $tag, 'error' => FALSE);
// check for tag type
if ($tag == 'login') {
$id = $_POST['id'];
$password = $_POST['password'];
$user_type = $POST['user'];
// test data
$response = array('tag' => $tag, 'error' => FALSE, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);
// check for user
$user = $db->getUserByIdAndPassword($user_type, $id, $password);
echo json_encode($response);
}
} else {
$response["error"] = TRUE;
$response["error_msg"] = "Required parameter 'tag' is missing!";
echo json_encode($response);
}
The query is run in this class, which is where I suspect I'm going wrong.
DB_Functions.php
class DB_Functions {
private $db;
// constructor
function __construct() {
require_once 'DB_Connect.php';
// connecting to database
$this->db = new DB_Connect();
$this->db->connect();
// destructor
function __destruct() {
}
/**
* Get user by id and password
*/
public function getUserByIdAndPassword($user_type, $id, $password) {
$t = 'T';
if ($user_type !== $t) {
$result = mysqli_query("SELECT * FROM smiths WHERE id = '$id'") or die(mysqli_error());
} else {
$result = mysqli_query("SELECT * FROM traders WHERE id = '$id'") or die(mysqli_error());
}
// check for result
$no_of_rows = mysqli_num_rows($result);
if ($no_of_rows > 0) {
$result = mysqli_fetch_array($result);
$retrieved_password = $result['password'];
// check for password equality
if ($retrieved_password == $password) {
return $result;
}
} else {
// user not found
return false;
}
}
}
And finally this class manages the msql connection, I think I need the $con from here for the previous mysqli_query? I'm not sure how to call it.
DB_Connect.php
class DB_Connect {
// constructor
function __construct() {
}
// destructor
function __destruct() {
// $this->close();
}
// Connecting to database
public function connect() {
require_once 'include/Config.php';
// connecting to mysql
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
// Check connection
if (!$con)
{
die("Connection error: " . mysqli_connect_error());
}
// selecting database
mysqli_select_db($con, DB_DATABASE) or die(mysqli_connect_error());
// return database handler
return $con;
}
// Closing database connection
public function close() {
mysqli_close();
}
}
Could anyone help set me in the right direction? The JSON response is coming back with the error;
W/System.err: org.json.JSONException: End of input at character 0
[EDIT]
Alright so I have taken the following lines from DB_Connect.php and put them into the method in DB_Functions.php.
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
mysqli_select_db($con, DB_DATABASE) or die(mysqli_connect_error());
This then allows me to fix the syntax of msqli_query as so;
$result = mysqli_query($con, "SELECT * FROM smiths WHERE id = '$id'") or die(mysqli_error());
This has fixed my issue, however hacky/messy it may seem.

Answer 1

I have taken the following lines from DB_Connect.php and put them into the method 'getUserByIdAndPassword' in DB_Functions.php.
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
mysqli_select_db($con, DB_DATABASE) or die(mysqli_connect_error());
This then allows me to fix the syntax of msqli_query as so;
$result = mysqli_query($con, "SELECT * FROM smiths WHERE id = '$id'") or die(mysqli_error())
And this kids is why you don't use deprecated code like some Frankenstein madman.