如何设置打开弹出窗口并在另一个页面中获取Chacked值

Question

1. （1）选择预订（更改值）然后打开弹出窗口即可。

2. （2）在Checked NetAmount中打开弹出窗口，然后我想在文本中获得Total 框。

3. （3）然后Cheked NetAmount在索引页面中显示此记录。

   我想点（2,3）。

的index.php

script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
$(function ()
{
$("#Debit_id").change(function ()
{ 
var Debit_id = document.getElementById('Debit_id').value;
if(Debit_id!="")
{
window.open("getpopup.php?Debit_id="+Debit_id,'popup','width=700,height=400,left=200,top=200,scrollbars=1');
}
});
});
</script>


Book : <select name="Debit_id" id="Debit_id">
<option value="" selected="selected">Select Book</option>
<?php 
$result1 = mysql_query("SELECT AccountName,CID FROM account")or die(mysql_error());         
while($row1 = mysql_fetch_array($result1))
if($row1)
{ ?>
<option value="<?php echo $row1['CID']; ?>"><?php echo  $row1['AccountName']; ?></option><?php 
} ?>
</select>
<br />
NetAmount Total : <input type="text" class="input" name="NetAmount" id="TotalCost">

popup.php

<?php
include("../config.php");
$Debit_id = intval($_GET['Debit_id']);
?>
<table>
<tr>
<th>SrNo</th>
<th>EntryDate</th>
<th>NetAmount</th>
</tr>
<?php 
$sql="SELECT * FROM transaction WHERE Credit_id = '".$Debit_id."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
$SrNo =$row['SrNo'];
$EntryDate = $row['EntryDate'];
$NetAmount =$row['NetAmount'];
?>
<tr>
<td><?php echo $SrNo ?></td>
<td><?php echo $EntryDate ?></td>
<td><?php echo $NetAmount ?><input type="checkbox"  name="<?php echo $SrNo ?>" value="<?php echo $NetAmount ?>"></td>
</tr>

<?php }?>
<tr>
<td colspan="3" align="center">
<input type="submit" value="ok" />
</td>
</tr>
</table>