Thursday, January 25, 2018

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





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



That 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


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