SQL LIKE query failing - fatal error in prepared statement

I have the following code:

$countQuery = "SELECT ARTICLE_NO FROM ? WHERE upper(ARTICLE_NAME) LIKE '% ? %'";
if ($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);
}

Which should work fine. However I receive the error that :

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? WHERE upper(ARTICLE_NAME) LIKE '%?%'' at line 1

If I put

SELECT ARTICLE_NO FROM AUCTIONS WHERE upper(ARTICLE_NAME) LIKE '% o %';

The query works fine. $table is defined above, and query is received from GET, and both are correct valid values. Why is this failing?

edit: changing to:

$countQuery = "SELECT ARTICLE_NO FROM AUCTIONS1 WHERE upper(ARTICLE_NAME) LIKE '% ? %'";

if ($numRecords = $con->prepare($countQuery)) {

    $numRecords->bind_param("s", $query);

results in the error:

Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in C:\Program Files\EasyPHP 3.0\www\prog\get_records.php on line 38

Commands out of sync; you can't run this command now

where as

$countQuery = "SELECT ARTICLE_NO FROM AUCTIONS1 WHERE upper(ARTICLE_NAME) LIKE ?";

if ($numRecords = $con->prepare($countQuery)) {

    $numRecords->bind_param("s", "%".$query."%");

results in

Fatal error: Cannot pass parameter 2 by reference in C:\Program Files\EasyPHP 3.0\www\prog\get_records.php on line 38

and lastly

$countQuery = "SELECT ARTICLE_NO FROM AUCTIONS1 WHERE upper(ARTICLE_NAME) LIKE ? ";

if ($numRecords = $con->prepare($countQuery)) {

    $numRecords->bind_param("s", $query);

will give only:

Commands out of sync; you can't run this command now

Is it impossible to use a paramter for a LIKE statament?

From stackoverflow

• 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);
  }
  

  Joshxtothe4 : nope, this did not solve it, or any similar tricks.

• Afaik you can't use placeholders for identifiers with mysqli and prepare statements. So you'd have to manually interpolate the tablename into the query.

  Joshxtothe4 : huh. They have failed before when I have not had the tablename as a parameter.

• For LIKE clause, use this:

  SELECT ARTICLE_NO FROM AUCTIONS1 WHERE upper(ARTICLE_NAME) LIKE CONCAT('%', ?, '%')
  

  As for the table name, it's an extremely bad practice to have table names as parameters.

  If for some reason you still need to do it, you'll need to embed it into the query text before preparing the query:

  $countQuery = "SELECT ARTICLE_NO FROM $table_name WHERE upper(ARTICLE_NAME) LIKE CONCAT('%', ? ,'%')";
  if ($numRecords = $con->prepare($countQuery)) {
      $numRecords->bind_param("s", $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);
  }
  

  Joshxtothe4 : Other people have said this, but passing the table name as a parameter works fine? Is it bad practice to do so?

  Quassnoi : Passing table name as a parameter won't work, MySQL won't be able too prepare such a statement. You cannot have a prepared statement with a table name as a parameter.

• Have you issued

  mysqli_free_result($result);
  

  after the last query? That's the command out of sync error.

  This should work however

  $countQuery = "SELECT ARTICLE_NO FROM AUCTIONS1 WHERE upper(ARTICLE_NAME) LIKE ?";
  if ($numRecords = $con->prepare($countQuery)) {
      $numRecords->bind_param("s", "%".$query."%");
  

  Wondering what is in the $query variable. Try doing this instead

  $query = '%'.$query.'%';
  $numRecords->bind_param("s", $query);
  

  Joshxtothe4 : I have no idea where to issue that..I have freed everything as far as I can see, and still get that error.

  michal kralik : This can happen, for example, if you are using mysql_use_result() and try to execute a new query before you have called mysql_free_result(). It can also happen if you try to execute two queries that return data without calling mysql_use_result() or mysql_store_result() in between.