php - displaying value of dynamic checkbox in next page . and inserting such value -

i've created dynamic checkbox, failed display value on page. work have done following:

index.php

<form method="post" action="print.php">     <?php      $host="localhost";     $username="root";     $password="";     $database="checkbox";      mysql_connect($host,$username,$password);      mysql_select_db("$database");     //create query     $sql = "select test, rate lab";     $result = mysql_query($sql) or die(mysql_error());     while($row = mysql_fetch_assoc($result)) {         echo <<<eol         <input type="checkbox" name="name[]"value="$row['test']}/{$row['rate']}"/>                   {$row['test']}-{$row['rate']}<br />          eol;     }      ?>     <br>      <input type="submit" name="submit" value="add" /> </form> 

i trying display value on secon page called print.php:

<?php print $_post['name']; ?> 

see what's happening in code:- naming check box in array. when submission in php receive array name :- name[] $_post['name'] return array in php. when use print method can print variable value . can't print array or object . if use print/echo method print array/object print type. print array can use print_r() method or can use var_dump() check in variable. can access array favourite way loop. more print-r , var_dump please follow manual link [php.net manual][1] http://www.php.net/manual/en/function.var-dump.php