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in SQL by (6.1k points)
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I am having some difficulty checking if a Facebook User_id already exists in my database (if it does not it should then accept the user as a new one and else simply load the canvas application). I was running it on my hosting server and there was no problem, but on my localhost, it gives me the below error:

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

Here is 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 have read, this has something to do with my query being wrong, but it has worked on my hosting provider so that can not be it.

1 Answer

0 votes
by (12.7k points)
edited by

The query is failing and it's returning false.

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

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

For more information, check it here.

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