Update hide field only when "close" drop down selected

PHP
1

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 :&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:


<?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.

2

When checking $_POST values, I tend to use something like this:


$myfield='';
if(isset($_POST['thefield']) && !empty($_POST['thefield'])){
     $myfield=trim($_POST['thefield']);
}

What you’re doing here is checking to see if the POST variable exists and checks to make sure that it is not an empty value (such as an empty string, null, or zero).

It can also be written as:

$myfield=(isset($_POST['thefield']) && !empty($_POST['thefield'])) ? trim($_POST['thefield']) : '';

Then, you will need to check the values to see if they are the ones you are looking for.

if($myfield=='myvalue'){
//Do stuff here
}

Also, you should not be using mysql_* functions, as they have been depreciated. Use mysqli or PDO.

3

You’re just hiding it from displaying but the input field is still there. You want to disable it. I think this should do it using jQuery.


                if (this.value == "Close") {
                    $('#Other').show().find( "input" ).prop( "disabled", false );
                } else {
                    $('#Other').hide().find( "input" ).prop( "disabled", true );
                }

This will find all input tags in the “other” tr and set disabled to either true or false.