I have this line of code:
$query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`");
its returning this error:
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result
full code:
$query = mysqli_query($connection, "SELECT `id`, `image`, `link`, `order` FROM `galleryImages` INNER JOIN `galleries` ON `galleryImages`.`galleryId` = `galleries`.`id`");
$results = array();
while($row = mysqli_fetch_assoc($query)){
$results[] = $row;
}
return $results;