javascript - How Do Set open popup And get Chacked value in another page -
(1) select book(onchange value) open popup ok.
(2)open popup in checked netamount want total in text box.
(3)then cheked netamount display record in index page.
i want point(2,3).
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 $result1 = mysql_query("select accountname,cid account")or die(mysql_error()); while($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 $sql="select * transaction credit_id = '".$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>
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