Update hide field only when "close" drop down selected

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jackgoddy123
Forum Newbie
Posts: 14
Joined: Sat Jan 11, 2014 8:10 am

Update hide field only when "close" drop down selected

Post by jackgoddy123 »

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:

Code: Select all

<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 :&nbsp; <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:

Code: Select all

<?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.
User avatar
Celauran
Moderator
Posts: 6427
Joined: Tue Nov 09, 2010 2:39 pm
Location: Montreal, Canada

Re: Update hide field only when "close" drop down selected

Post by Celauran »

The value of $_POST['lead_close'] is being generated when the form is rendered. It will always have a value. You need to check the value of status to decide whether or not to include $lead_close in your query.
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