PHP script to check If user already exists in MySQL database - php

Question

I'm creating simple game for Facebook. All users who used app are written to database. I need always check If user already exists Is in database, how to do that correctly?
So I have variable $name = $user_profile['name']; It successfully returns user's name
And this is my part of code to check If user already exists in database.
$user_profile = $facebook->api('/me');
$name = $user_profile['name'];
$mysqli = new mysqli("host","asd","pw","asdf");
echo "1";
$sql = "SELECT COUNT(*) AS num FROM myTable WHERE userName = ?";
echo "2";
if ($stmt = $mysqli->prepare($sql)) {
echo "3";
$stmt->bind_param('s', $name);
echo "4";
$stmt->execute();
echo "5";
$results = $stmt->get_result();
echo "6";
$data = mysqli_fetch_assoc($results);
echo "7";
}
if($data['num'] != 0)
{
echo "bad";
print "user already exists\n";
} else {
echo "good";
$apiResponse = $facebook->api('/me/feed', 'POST', $post_data);
print "No user in database\n";
}
}
This code not working, It should post data on user's wall If user not exists in database. I spent many time to find reason why, but unsuccessfully. After debugging It don't show any errors. To find which line is incorrect after every line I used echo "number" so now I know which line is incorrect. It prints 1 2 3 4 5 and stucks. (everything what are below the code not loading.) So that means this line $results = $stmt->get_result(); is incorrect. But I misunderstood what's wrong with this line?
If I comment this line all code loading (then print 1 2 3 4 5 6 7 No user in database! and It post data on user's wall.) but in this case program always do the same, not checking database.
Also I've tried to change COUNT(*) to COUNT(userName), but the same.
So could you help me, please?
I've read this: Best way to check for existing user in mySQL database? but It not helped me.
P.s. In this case i need to use FB username.

Answer 1

Can you try this, $stmt->fetch() instead of mysqli_fetch_assoc($results)
$mysqli = new mysqli("host","asd","pw","asdf");
echo "1";
/* Create the prepared statement */
$stmt = $mysqli->prepare("SELECT COUNT(*) AS num FROM myTable WHERE userName = ?") or die("Prepared Statement Error: %s\n". $mysqli->error);
/* Execute the prepared Statement */
$stmt->execute();
/* Bind results to variables */
$stmt->bind_result($name);
$data = $stmt->fetch();
if($data['num'] > 0)
{
echo "bad";
print "user already exists\n";
} else {
echo "good";
$apiResponse = $facebook->api('/me/feed', 'POST', $post_data);
print "No user in database\n";
}
/* Close the statement */
$stmt->close();
Ref: http://forum.codecall.net/topic/44392-php-5-mysqli-prepared-statements/