mysqli fetch array expects parameter 1 to be mysqli result boolean given in

0 votes

I am having an issue with the function where it checks if a Facebook user id is there in the database or not. It works in the hosting server but gives an error in the local host.

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

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;
}

Can someone help me solve this issue?

Apr 30, 2022 in Other DevOps Questions by Kichu
• 19,050 points
485 views

1 answer to this question.

0 votes

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

To solve this just put this after mysqli_query() :

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

For more information: http://www.php.net/manual/en/mysqli.error.php.

answered May 1, 2022 by narikkadan
• 53,920 points

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