ajax和jquery使用php和mysql选择box post方法

Question

我正在使用ajax和jquery选择框发布脚本。

我的问题是价值ID。当我clik post，然后数据库表只显示id号。但我想发布is_kategorisi。我能做什么 ？ 这是我的选择框：

 <?php
    include('db.php');
    $is_kategorisi =$_POST['id'];
    $query = "SELECT * FROM business_category";
    $res   = mysql_query($query);
    ?>
    <select id="parent_category" name="sirket_kategorisi" accept-charset="UTF-8" />
      <?php
      echo "<option value=''>Şirket Kategoriniz</option>";
      while($row = mysql_fetch_array($res))
      {
        echo "<option value='".$row['is_kategorisi']."'>".ucfirst($row['is_kategorisi'])."</option>";
      }
      ?>
    </select>

这是我发布的php代码：

<?php 
  include("includes/connect.php");

     if(isset($_POST['submit'])){
         $sirket_kategorisi = $_POST['sirket_kategorisi'];
$query ="INSERT INTO `users` (`sirket_kategorisi`) VALUES ('$sirket_kategorisi')";   
           $result = mysql_query($query) or die(mysql_error());

           if($result){
               echo("<center><h1>OK!</h1></center>");      
               }
           else {      
               echo("<center><h1>Not OK!</h1></center>");      
               }
       }
?> 

ajax代码：

<script type="text/javascript">
$(document).ready(function()
{
  $("#parent_category").change(function()
  {
    var category_id = $(this).val();
    if(category_id != '')  
     {
      $.ajax
      ({
         type: "POST",
         url: "get_child.php",
         data: "category_id="+ category_id,
         success: function(option)
         {
           $("#child_category").html(option);
         }
      });
     }
     else
     {
       $("#child_category").html("<option value=''>-- No category selected --</option>");
     }
    return false;
  });
});
</script>

get_child.php

<?php
include('db.php');
?>

<?php
if(isset($_POST['category_id']) && $_POST['category_id'] != '')
{
  $category_id = $_POST['category_id'];
  $category_id = mysql_real_escape_string($category_id);
  $query = "SELECT * FROM business_category_name where catg='".$category_id."'";
  $res = mysql_query($query);
  if(mysql_num_rows($res))
  {
    while($row = mysql_fetch_array($res))
    {
      echo "<option value='".$row['category_name']."'>".ucfirst($row['category_name'])."</option>";
    }
  }
}


?>