Populate another drop down list with data from database and insert into another datab

PHP
1

Hello experts,
I have created a dropdown list on each click of drop down another drop down list appears from drop1 database. Below is the code of the same:

dropdown1.php


<html>
<head>
<script>
function showUser(str)
{
if (str=="")
  {
  document.getElementById("txtHint").innerHTML="";
  return;
  }
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","drop2.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a plan:</option>
<option value="1">Dedicated</option>
<option value="2">VPS</option>

</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here.</b></div>

</body>
</html>

drop2.php:


<?php
$q = intval($_GET['q']);

$con = mysqli_connect('localhost','root','','test');
if (!$con)
  {
  die('Could not connect: ' . mysqli_error($con));
  }

mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM drop WHERE id = '".$q."'";

$result = mysqli_query($con,$sql);


<tr>

<th>plan1</th>
<th>plan2</th>

</tr>";*/
  echo "Plan";
  echo "<select>";

while($row = mysqli_fetch_array($result))
  {

   echo "<option>" . $row['plan1'] . "</option>";
   echo "<option>" . $row['plan2'] . "</option>";

  }
//echo "</table>";
echo "</select>";
mysqli_close($con);
?>

The above code works perfectly fine for shown the proper dropdown.
Now i want to insert the second dropdown value in another database.
And i am quite stuck in that. :frowning:

2