php - mysqli prepared statement doesn't seem to work -

i have following code:

<?php if(!empty($_post)){     $db = new mysqli('localhost', 'root', '', 'verif1');     $password = $_post['pass'];     $username = $_post['user'];         $table = "admins";     $password = hash("sha256", $password);     $statement = $db->prepare("select count(*) {$table} username = ? , password = ?");     $statement->bind_param("ss", $username, $password);     $statement->execute();     $statement->bind_result($numrows);     if($numrows == 1){         echo "yeah correct!<br>";     }else{         echo "no :(<br>";     } }   ?>  <form action="" method="post">     <input type="textbox" name="user">     <br>     <input type="password" name="pass">         <br>     <input type="submit"> </form> 

i'm sure correct (no error throwing, nothing) doesn't work! tried print_r() see objected has returned , this: after execute() this:

print_r($statement); die(); 

and gives me this:

mysqli_stmt object ( [affected_rows] => -1 [insert_id] => 0 [num_rows] => 0 [param_count] => 2 [field_count] => 1 [errno] => 0 [error] => [error_list] => array ( ) [sqlstate] => 00000 [id] => 1 )

if check documentation - http://www.php.net/manual/en/mysqli-stmt.bind-result.php see you've forgotten fetch() data

$statement->fetch();