mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in [duplicate]

Question

I'm have some trouble checking if a Facebook User_id already exists in my database (if it doesn't it should then accept the user as a new one and else just load the canvas application). I ran it on my hosting server and there was no problem, but on my localhost it gives me the following error:

mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in

Here's my code:

<? $fb_id = $user_profile['id']; $locale = $user_profile['locale'];  if ($locale == "nl_NL") {     // Checking User Data @ WT-Database     $check1_task = "SELECT * FROM `users` WHERE `fb_id` = " . $fb_id . " LIMIT 0, 30 ";     $check1_res = mysqli_query($con, $check1_task);     $checken2 = mysqli_fetch_array($check1_res);     print $checken2;     // If the user does not exist @ WT-Database -> insert     if (!($checken2)) {         $add = "INSERT INTO users (fb_id, full_name, first_name, last_name, email) VALUES ('$fb_id', '$full_name', '$first_name', '$last_name', '$email')";         mysqli_query($con, $add);     }     // Double-check, the user won't be able to load the app on failure inserting to the database     if (!($checken2)) {         echo "Excuse us " . $first_name . ". Something went terribly wrong! Please try again later!";         exit;     } } else {     include ('sorrylocale.html');     exit; } 

I've read it has something to do with my query being wrong, but it has worked on my hosting provider so that can't be it!

Answer 1

The query given to mysqli_query() is failing and returning false.

Put this after mysqli_query() to see what's going on.

if (!$check1_res) {     printf("Error: %s\n", mysqli_error($con));     exit(); } 

For more information:

http://www.php.net/manual/en/mysqli.error.php