calling javascript function from php code

by **sheeshkamal** » Thu Aug 30, 2012 9:07 pm

Hi All,

I have problem in follwing script.

<?php
if(isset($_GET['submit'])){
if(!empty($_GET['name'])){
$name=$_GET['name'];

echo'<script type="text/javascript">show();</script>';

}
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
#display{ display:none; border:2px dashed #0033FF; width:200px}
</style>
<script type="text/javascript">
function show(){

document.getElementById("display").style.display="block";
}
</script>
</head>

<body>
<form method="get" action="form.php">
<input type="text" name="name" />
<input type="submit" name="submit" />

</form>
<div id="display">

<?php echo $name; ?>
</div>
</body>
</html>

I am trying to make visible div when php code executes.

Can Anyone please help me on this????

by **NigelRen** » Fri Aug 31, 2012 11:27 pm

You cannot call Javascript from PHP, PHP is only executed on the server, whilst Javascript is a client side language.
What you can do with PHP is to create the HTML that reflects what you want, so if you had...

Code: Select all: if(isset($_GET['submit'])){ if(!empty($_GET['name'])){ $name=$_GET['name']; ?> <div id="display"> <?php echo $name; ?> </div> <?php } }

Then this part of the html will only be included only when you have the fields set.

calling javascript function from php code

calling javascript function from php code

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