Update hide field only when "close" drop down selected

jackgoddy123 · January 26, 2014

Hello to all,

I am updating a dropdown option field. But when i choose "Close" option then a textbox field with the current date appears.

Below is the code for it:

<html>
<head>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>

<script>
$(document).ready(function () {
            $('#status').change(function () {
                if (this.value == "Close") {
                    $('#Other').show();
                } else {
                    $('#Other').hide();
                }

            });
        });


</script>
</head>
<body>
<form action="status2.php" method="post">
Status
<select id="status" name="status" value="<?=$row['status']?>">
                        <option>Choose</option>
                        <option value="Open">Open</option>
                        <option value="Inprocess">Inprocess</option>
                        <option value="Close">Close</option>
                    </select>
  <table>
             <tr id="Other" style="display: none">
                <td>
                   <!-- <input id="txtOthers" type="text" /> -->
				   <label>Lead Close Date :  <input type="text"  name="lead_close" value="<?php $b = time ();
  print date("d-m-y",$b) ; 
 ?> "  ><br />
  </td>
            </tr>
        </table>

<br/>
<input type="submit" value="Update" name="submit">
</form>
</body>
</html>

The above code works perfectly fine.

But when i update the records then even the hide textbox(lead_close) gets update. It must only update when the close option gets selected orelse only status with "Open" or "Inprocess" must update.

Below is my update code:

<?php
$konek = mysql_connect("localhost","root","") or die("Cannot connect to server");
mysql_select_db("test",$konek) or die("Cannot connect to the database");

$status= ($_POST['status'])?$_POST['status']:'';

$lead_close= ($_POST['lead_close'])?$_POST['lead_close']:'';

mysql_query("update public set status='".$status."', lead_close='".$lead_close."' where id='1'");
?>

I dont know where i am missing it.

Just need some correction in my code.

Thanks in advance.

Psycho · January 26, 2014

Just change the logic to your processing code. If the status does not equal close, then don't update the close date. Also, the code for setting the status has a flaw since it would allow any value - not just the value you have on the form. You need to validate ALL date being sent from a user.

$konek = mysql_connect("localhost","root","") or die("Cannot connect to server");
mysql_select_db("test",$konek) or die("Cannot connect to the database");

$status = ($_POST['status']) ? $_POST['status'] : '';
$lead_close = ($status=='Close' && $_POST['lead_close']) ? $_POST['lead_close'] : '';

$query = "UPDATE public SET status='{$status}', lead_close='{$lead_close}' WHERE id='1'";
mysql_query($query);

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Update hide field only when "close" drop down selected

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