为什么数据没有通过javascript传递给php

i am using the following form:

<form id="dataForm" method="post">
  <h2 id="formheader"> Update Product Description</h2>
    <div>
      <label>Product Name:</label>
      <input class="inputForm" id="orginalName" type="text" name="Name">
    </div>
    <div>
      <label>New Description:</label>
      <input class="inputForm" id="newDescription" type="text" name="newDescription">
    </div>
    <div id="theSubmit">
      <button id="editDescription">Submit</button>
    </div>
  </form>

and using the following simple php, which when used with action=editProductDes.php works...

 $Name = $_POST['Name'];
 $Description = $_POST['newDescription'];

 if($Name !="" && $Description !=""){
 $sql = "UPDATE PRODUCTS SET P_Description = '$Description' WHERE P_NAME = '$Name'";
 $conn->exec($sql); 

and then when i use the following java script the data is not passed through and I cannot see why as I have a similar function and form where the JavaScript works fine, can anyone see why the data is not passing through?

function editDescription(){
    xmlhttp = new XMLHttpRequest();
    var name = document.getElementById("orginalName");
    var Description = document.getElementById("newDescription");

    var data_seen = false;
        // this is a flag to record whether any data has been seen. Used in the guard ofthe alert statement.
    if (name.value !="" && Description.value !="" ){
        data_seen = true;
        xmlhttp.open("POST","editDescription.PHP",true);
        xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        xmlhttp.send("Name=" + name.value + "&Description=" + Description.value);
    }
    if (!data_seen) {
        alert("please enter some data");
    }
   }

submitButton = document.getElementById("editDescription");
submitButton.addEventListener("click", editDescription);

You are posting to editDescription.PHP instead of editProductDes.php

Change the following:

xmlhttp.open("POST","editProductDes.php",true);

You are also sending the data in your post under another name than you expect it to be in your PHP code (Description instead of newDescription) - change:

xmlhttp.send("Name=" + name.value + "&newDescription=" + Description.value);