【发布时间】:2015-02-28 10:02:07
【问题描述】:
我正在尝试将我单击的
代码:
offices.php
<?php
include 'connect.php';
$page = 'offices';
include 'header.php';
if (isset($_POST['postdata'])){
$filter1 = $_POST['postdata'];
echo $filter1;
}
?>
<!DOCTYPE html>
<html>
<head>
<script src="js/jquery.min.js"></script>
</head>
<body>
<div class="filter1">
<ul>
<li class="fteacher" data-value="teacher">Professor</li>
<li class="foffice" data-value="office">Gabinete</li>
</ul>
</div>
<script src="js/offices.js"></script>
</body>
</html>
offices.js
$('.filter1 li').click(function(){
$.ajax({
type: 'POST',
url: 'offices.php',
data: {'postdata': $(this).attr('data-value')},
success: function(msg){
alert('Success');
}
});
});
【问题讨论】:
-
为什么你认为 PHP 没有被执行?