OK, I've got a drop-down that displays contents from the MySQL DB properly in a table (table is ugly, but I'm working on the layout). Here's the HTML of the page that shows the drop-down and table (there's 48 entries, I've truncated it to 5):
Code:
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script type="text/javascript" src="selectuser.js"></script>
</head>
<body>
<form>
<span class="headingText">Select the name: </span>
<select name="users" class="formText" onChange="showUser(this.value)">
    <option selected="selected">Select...</option>
    <option value="1">A1</option>
    <option value="2">B2</option>
    <option value="3">C3</option>
    <option value="4">D4</option>
    <option value="5">E5</option>
</select>
<br />
</form>
<div class="contentText" id="txtHint">Site information will be listed here.</div>
</body>
</html>
And the selectuser.js code:
Code:
var xmlhttp;

function showUser(str)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
  {
  alert ("Browser does not support HTTP Request");
  return;
  }
var url="getuser.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}

function stateChanged()
{
if (xmlhttp.readyState==4)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}

function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
  {
  // code for IE7+, Firefox, Chrome, Opera, Safari
  return new XMLHttpRequest();
  }
if (window.ActiveXObject)
  {
  // code for IE6, IE5
  return new ActiveXObject("Microsoft.XMLHTTP");
  }
return null;
}
And the getuser.php code:
PHP Code:
<?php

$q
=$_GET["q"];

$hostname="SERVER";
$username="USERNAME";
$password="PASSWORD";
$dbname="DBNAME";
$usertable="TABLENAME";

mysql_connect($hostname,$username$password) or die ("<html><script language='JavaScript'>alert('Unable to connect to database! Please try again later.'),history.go(-1)</script></html>");
mysql_select_db($dbname);

$query "SELECT * FROM $usertable WHERE id = $q";

$result mysql_query($query);

echo 
"<table border='1'>
<tr>
<th>Name</th>
<th>RunBy</th>
<th>Director</th>
<th>Address</th>
<th>City</th>
<th>District</th>
<th>Province</th>
<th>Telephone</th>
<th>LatLong</th>
<th>GeoNotes</th>
<th>Comments</th>
</tr>"
;

while(
$row mysql_fetch_array($result))
  {
  echo 
"<tr>";
  echo 
"<td>" $row['Name'] . "</td>";
  echo 
"<td>" $row['RunBy'] . "</td>";
  echo 
"<td>" $row['Director'] . "</td>";
  echo 
"<td>" $row['Address'] . "</td>";
  echo 
"<td>" $row['City'] . "</td>";
  echo 
"<td>" $row['District'] . "</td>";
  echo 
"<td>" $row['Province'] . "</td>";
  echo 
"<td>" $row['Telephone'] . "</td>";
  echo 
"<td>" $row['LatLong'] . "</td>";
  echo 
"<td>" $row['GeoNotes'] . "</td>";
  echo 
"<td>" $row['Comments'] . "</td>";
  echo 
"</tr>";
  }
echo 
"</table>";
?>
Minus the output table formatting which I'm still learning how to fix, it works swimmingly. However, I want to know if this is possible:

The information that's dumped into the LatLong cell, is there any way to make that same info auto-populate another text field in another form on the same page (id="flyhere") once it shows up in the table? And re-populate it once the selection in the dropdown menu is changed?

Or make the cell in that table clickable, so if info does populate that cell, you can click on it and have it populate the other "flyhere" text box in the other form?