I want to incorporate a session variable $_SESSION['LI']
into the jQuery with an if statement.
So if($_SESSION['LI'] == 'falsus')
hide the #feedback else show #feedback.
Thanks
<?php
session_start();
$_SESSION['LI'] = 'falsus';
?><html>
<head>
<title></title>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#feedback').hide();
});
</script>
</head>
<body>
<div id="feedback">Hello</div>
<? print_r($_SESSION);?>
&l开发者_JS百科t;/body>
</html>
Could you just do something like this:
<?php
session_start();
$_SESSION['LI'] = 'falsus';
?>
<html>
<head>
<title></title>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.2/jquery.min.js"></script>
<?php if($_SESSION['LI'] == 'falsus'): ?>
<script type="text/javascript">
$(document).ready(function() {
$('#feedback').hide();
});
</script>
<?php endif; ?>
</head>
<body>
<div id="feedback">Hello</div>
<? print_r($_SESSION);?>
</body>
</html>
This will only add the script to hide the #feedback div when the session variable is "falsus"
. If it is NOT "falsus"
, then the whole script block is omitted.
<?php
session_start();
$_SESSION['LI'] = 'falsus';
?><html>
<head>
<title></title>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
<?php if ($_SESSION['LI'] == 'falsus') { ?>
$('#feedback').hide();
<?php } else { ?>
$('#feedback').show();
<?php } ?>
});
</script>
</head>
<body>
<div id="feedback">Hello</div>
<? print_r($_SESSION);?>
</body>
</html>
You can weave your code with php wherever you want.
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