PHP script throwing 500 internal error - php

Question

I have a Javascript script that sends an Ajax request via POST to the following PHP script. $db is passed from up above, but I know that part works. I'm getting a 500 Internal Server Error from this. I am unable to see the server logs, so I can't get more detailed information than this. Anyone know what I can do to fix it?
function register($db) {
$user = $_POST['username'];
$pass = $_POST['password'];
$query = "SELECT username FROM users WHERE username='$user'";
$result = mysqli_query($db, $query);
if (mysqli_fetch_array($result)[0] == $user) {
echo "taken";
}
else {
$query = "INSERT INTO users (username, password) VALUES ('$user', '$pass')";
if (mysqli_query($db, $query)) {
echo "success";
}
}
}

Answer 1

This is more of a corollary than an actual answer, but it is far too long for the format of a comment.
For the sake of a followup to #MarcB's comment, you should change your code to be invulnerable to SQL injection:
$user = $_POST['username'];
$pass = $_POST['password'];
$link = new mysqli($host, $username, $password, $database);
if(!$link)
{
die("Unable to connect to MySQL");
}
$stmt = $link->prepare("SELECT username FROM users WHERE username=?");
$stmt->bind_param("s", $user);
$stmt->bind_result($testUsername);
$stmt->execute();
$stmt->store_result();
$stmt->fetch();
if($stmt->num_rows > 0)
{
echo "taken";
}
else
{
$stmt = $link->prepare("INSERT INTO users (username, password) VALUES(?, ?)");
$stmt->bind_param("ss", $user, $pass);
$stmt->execute();
echo "success";
}

Answer 2

I believe you have a parse error which is fixed in later versions of PHP. The following line of code is not valid <= PHP 5.4:
if (mysqli_fetch_array($result)[0] == $user) {
In order to fix this, you need to change that line to the following:
$row = mysqli_fetch_array($result);
if ($row[0] == $user) {
Also, your code also has a pretty severe SQL injection vulnerability. You need to sanitise your variables before using them in queries. For integers and floats, I recommend you do the following:
$myInt = (isset($_POST['myInt']) ? (int)$_POST['myInt'] : null);
$myFloat = (isset($_POST['myFloat']) ? (float)$_POST['myFloat'] : null);
For strings, you should use mysqli_real_escape_string() as follows:
$myStr = (isset($_POST['myStr']) ? mysqli_real_escape_string($_POST['myStr']) : null);
In your case, your $user and $pass variables need to be sanitised as the string is above. The reason for forcing the types of numbers you're receiving via $_POST is that it entirely negates any possibility of someone passing through a value which is not valid. Anything which is not 0, 1, 2, 3, 4, etc that is cast as (int) will become 0.