sql server - php sql odbc_execute() with LIKE not working -

so have piece of code not returning (the echo returns nothing , should returning 2 rows):

<?php  include "connection.php";  $cliente = $_post["cliente"];  $select = "select cliente, nomcli clix1 nomcli ? order nomcli";  $stmt = odbc_prepare($con, $select);  //preparing array parameter $prep_array = array();  $prep_array[] = "'%$cliente%'";  $rs = odbc_execute($stmt, $prep_array);  $nombres = array();  $clienteids = array();  //if prepare statement successful if($rs) {     $i = 0;      while($row=odbc_fetch_array($stmt))      {         $cliente_id = trim($row["cliente"]);          $nombre = utf8_encode(trim($row["nomcli"]));          $nombres[$i] = $nombre;          $clienteids[$i] = $cliente_id;          $i++;      }      echo json_encode($nombres) . "|" . json_encode($clienteids); }  else {     echo "error"; }  odbc_close($con);   ?> 

i know problem not parameter pass on odbc_execute() because if this, doesn't return anything(with %mich% should display 2 rows):

$rs = odbc_execute($stmt, array("%mich%")); 

do see wrong in code?

please let me know , in advance.

update ------

i made changes on code suggested on answer below , getting new error now:

warning: odbc_execute(): can't open file %mich% 

where mich text entered search on database.

i found following may relate: odbc prepared statements in php

$prep_array = array(); $prep_array[] = "'%$cliente%'"; $rs = odbc_execute($stmt, $prep_array); 

i think double quotes might causing issue.