SQL LIKE query failing - fatal error in prepared statement

Question

I have the following code:

$countQuery = \"SELECT ARTICLE_NO FROM ? WHERE upper(ARTICLE_NAME) LIKE \'% ? %\'\";
if ($numRecords = $con->prepare($countQuer         


        

Answer 1

Try the following instead:

$countQuery = "SELECT ARTICLE_NO FROM ? WHERE upper(ARTICLE_NAME) LIKE ?";
if ($numRecords = $con->prepare($countQuery)) {
    $numRecords->bind_param("ss", $table, "%$brand%");
    $numRecords->execute();
    $data = $con->query($countQuery) or die(print_r($con->error));
    $rowcount = mysql_num_rows($data);
    $rows = getRowsByArticleSearch($query, $table, $max);
    $last = ceil($rowcount/$page_rows);
}