首页 > 代码库 > AJAX 方式
AJAX 方式
<!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=gb2312" />
<title>Untitled Document</title>
<script src="http://www.mamicode.com/js/jquery.js"></script>
<script language="javascript">
function checkemail(){
if($(‘#email‘).val() == ""){
$(‘#msg‘).html("please enter the email!");
$(‘#email‘).focus;
return false;
}
if($(‘#address‘).val() == ""){
$(‘#msg‘).html("please enter the address!");
$(‘#address‘).focus;
return false;
}
ajax_post();
}
function ajax_post(){
$.post("test2.php",{email:$(‘#email‘).val(),address:$(‘#address‘).val()},
function(data){
//$(‘#msg‘).html("please enter the email!");
//alert(data);
$(‘#msg‘).html(data);
},
"text");//这里返回的类型有:json,html,xml,text
}
</script>
</head>
<body>
<form id="ajaxform" name="ajaxform" method="post" action="action.php">
<p>
email<input type="text" name="email" id="email"/>
</p>
<p>
address<input type="text" name="address" id="address"/>
</p>
<p id="msg"></p>
<p>
<input name="Submit" type="button" value="http://www.mamicode.com/submit" onclick="return checkemail()"/>
</p>
</form>
</body>
</html>
<?php
$email = $_POST["email"];
$address = $_POST["address"];
//echo $email;
//echo $address;
echo "success";
?>
AJAX 方式
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。