对于ajax请求
1.在您的网页中包含Jquery库。例如:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
2.点击按钮调用功能
<button type="button" onclick="create()">Click Me</button>
3.点击按钮,调用Javascript创建函数。
<script> function create () { $.ajax({ url:"test.php", //the page containing php script type: "post", //request type, dataType: 'json',data: {registration: "success", name: "xyz", email: "abc@gmail.com"} success:function(result){ console.log(result.abc);} }); }<script>在服务器端的test.php文件中,应读取action POST参数和相应的值,并以php格式执行操作并以json格式返回,例如
$regstration = $_POST['registration'];$name= $_POST['name'];$email= $_POST['email'];if ($registration == "success"){ // some action goes here under php echo json_enpre(array("abc"=>'successfuly registered'));}


