Question

Now I have convert my query with PDO format. But I must implement the query with old format for little next time. The code is like this:

function mysql_Select($sql) {
    $data=array();
    $params=func_get_args();
    $s=PreparaSQL($sql, $params);
    $res = mysql_query($s);
    if ($res && mysql_num_rows($res)){
        while( $dt = mysql_fetch_object($res)){
            $data[]=$dt;
        }
        return $data;
    }
}

function PreparaSQL($sql, $array_param){
   unset($array_param[0]);
   foreach ($array_param as $k => $v){
      $array_param[$k]=mysql_real_escape_string($v);       }
  return vsprintf( str_replace("params","%s",$sql), $array_param );  
}

And execute the function is :

$data=mysql_Select('SELECT concat(id," | ",wh2) as label,Id as kode,wh2 as nama,wh2 as value FROM wh011 where wh2 like %s',$_GET['where']);
echo json_encode($data);
flush();

And output have error with null value. I think the trouble is in this code:

return vsprintf( str_replace("params","%s",$sql), $array_param ); 

I really don’t know where is the error point.
Thank’s for you answer.

Answer 1

I think you have parameter 1 and 2 the wrong way around.

Try:

$sql = ” here is my string params did you find it?\n”;

echo vsprintf(str_replace(“params”,”%s”,$sql), “BOO!”);

Output

 here is my string BOO! did you find it?

I’d suggest just not doing the str_replace… ?