Ajax Php show value with change in dropdown what am i doing wrong?

programming
1

I am trying to make this.

  1. Suggest name in select menu as typed in text box
  2. Once the preferred name is clicked in the menu then show that users profile in other form (currently div would be ok).

Part 1 works good. But part 2 doesnot show any change. below are the codes.

<select name="names" id ="names" onchange="getdata();">
        <?php 
		while ($row = mysql_fetch_array ($result)) {
		?>
        <option id="names1" onchange="getdata()" value="<?php echo $row['name'];?>">
        <?php 
		echo $row['name'];
		?>
        </option>
        <?php
		$str = "";
            $i++;
        }
		?>
        </select><div id="ajaxDiv1">
	<?php include("find-name.php"); ?></div>

the getdata() is in javascript file which has code like


function getdata()
{
  var req = getXMLHTTP();
  if (req)
  {
        //function to be called when state is changed
        req.onreadystatechange = function()
        {
          //when state is completed i.e 4
          if (req.readyState == 4)
          {
                
				var ajaxSearchResults1 = document.getElementById("ajaxDiv1");
            	ajaxSearchResults1.innerHTML = req.responseText;
				// only if http status is "OK"
                if (req.status == 200)
                {
                        var new1 = document.getElementById('names').value;
						 var queryString1 = "?names=" + new1;
                }
                else
                {
                        alert("There was a problem while using XMLHTTP:
" + req.statusText);
                }
          }
        }
        req.open("GET", "find-name.php" + querystring1, true);
        req.send(null);
  }
}

and in the find-name.php i am for now only displaying the value recieved.


$names=$_GET['names'];
echo $names;

Where am i doing wrong Please help asap