PHP Variable not acting as a reference - php

Question

<?php
class User {
public $id;
public $counter;
public $removed;
}
$dB = json_decode(file_get_contents('dataBase.json'), true);
$dataBase = &$dB['noob'];
$userInDB = null;
$user = array('id' => (int)$_GET['id'], 'counter' => (int)$_GET['counter'], 'removed' => (bool)$_GET['removed']);
foreach ($dataBase as $usr) {
if ($usr['id'] == $user['id']) {
$userInDB = &$usr;
break;
}
}
if ($userInDB) {
$userInDB['counter'] = $userInDB['counter'] + $user['counter'];
$userInDB['removed'] = $user['removed'];
print_r($userInDB);
} else {
$dataBase[] = $user;
print_r($dataBase);
}
if(isset($_GET['id'])) {
$json = json_encode($user);
$updateddB = json_encode($dB);
file_put_contents('dataBase.json', $updateddB);
}
?>
Everything works except the part where I attempt to edit a value within an array. $userInDB is changed, but the section that it refers to within $dB isn't, even though I'm pretty sure I referred to it. Someone please help, I've had my head in knots.

Answer 1

You have the following loop:
foreach ($dataBase as $usr) {
if ($usr['id'] == $user['id']) {
$userInDB = &$usr;
break;
}
}
The $userInDB is a reference to $usr, however $usr was just created by the foreach loop, and is is not referenced to the original array it is looping over. So say, for example, you have this very simple case:
foreach ($foo as $var) {
$var++;
}
This does NOT affect $foo at all.
What you need to do is reference the $dataBase variable directly:
foreach ($dataBase as $key => $usr) {
if ($usr['id'] == $user['id']) {
$userInDB = &$dataBase[$key];
break;
}
}