Incomprehensible work of the script. Why does mysqli_query() return an empty object?

Question

PHP

Good afternoon.
I need your advice on how the script works.

$city =  iconv('windows-1251', 'UTF-8', $_GET['city']);
    $departament =  iconv('windows-1251', 'UTF-8', $_GET['departament']);
    $host = *************;
    $user = *************;
    $pass = *************;
    $data_base = **********;
    $connect = mysqli_connect($host,$user,$pass,$data_base) or die('Could not connect: ' . mysqli_error($connect));
    $sql="SELECT  * FROM 'имя таблицы' WHERE city = '".$city."'OR departament = '".$departament."'";
    $result = mysqli_query($connect,$sql) or die("ERROR: ".mysqli_error($connect));
    while($row = mysqli_fetch_array($result)){
        echo<<<"OUTPUT_DATA"
            <a href= $row[href] class="sj-j" data-r="34" data-c="9 ">
                <em><b>$row[vacancy]</b></em>
                <div></div>
                <i>$row[city]</i>
            </a>
OUTPUT_DATA;
    }
    mysqli_close($connect);

Task : there is a database with a table of vacancies (vacancy, requirements, city .. etc.). You need to make a form on the site, something similar to www.work.ua/jobs/by-company/297373/ (under the company description, "open vacancies").
How I did it: Through ajax - the request I get the necessary parameter (city or section) and then I process it all in the php script (the code is given above). When developing and debugging on OpenServer, everything worked. e. I choose the required city or section and, in fact, everything .. shows nothing.
I checked the login, password, host and database name 100 times, everything was entered correctly. There are no spaces after echo<<<"OUTPUT_DATA" either.
It is not clear why var_dump($result); returns object(mysqli_result)#2 (0) { } ?
Could it be related to the php version? (on site 5.2.) on openserver (5.3).
Tell me, please, what is my mistake?

Answer 1

WHERE city = '".$city."'OR
Where is the space before OR ?

Answer 2

one.
2.

while($row = mysqli_fetch_array($result)){
$did[] = $row;
}
var_dump($did);

And see what var_dump says

Answer 3

$mysqli = new mysqli($host, $user, $pass, $data_base);

$vacancies = $mysqli->prepare("SELECT * FROM 'имя таблицы' WHERE city = ? OR departament = ?");
        if ($vacancies) {
            $vacancies->bind_param('ss', $city, $departament); 
            $vacancies->execute();
            $vacancies->store_result();
            $countRows = $vacancies->num_rows;
            if ($countRows != 0){
            //на каждый столбец нужно забиндить по переменной, либо указывать в SELECT конкретные и биндить конкретные.
              $vacancies->bind_result($href, $vacancy, $city_result, $departament_result); 
                while ($vacancies->fetch()) {
                      echo '<a href="' . $href . '" class="sj-j" data-r="34" data-c="9 ">
                                <em><b>' . $vacancy . '</b></em>
                                <div></div>
                                <i>' . $city_result . '</i>
                            </a>';
                }
            }
            else{
                echo 'По вашему запросу ничего не найдено';
            }
        }