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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=php]
<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:
[code=php]
<?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.