AJAX PHP无刷新form表单提交_H5之家 - 中国HTML5教程资源分享第一站

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AJAX PHP无刷新form表单提交，有需要的朋友可以参考下。

ajax.php:

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

//获取表单对象和用户信息值
var f = document.user_info;
var userName = f.user_name.value;
var userAge = f.user_age.value;
var userSex = f.user_sex.value;

//接收表单的URL地址
var url = "./ajax_output.php";

//需要POST的值，把每个变量都通过&来联接
var postStr = "user_name="+ userName +"&user_age="+ userAge +"&user_sex="+ userSex;

//实例化Ajax
//var ajax = InitAjax();

//通过Post方式打开连接
ajax.open("POST", url, true);

//定义传输的文件HTTP头信息
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");

//发送POST数据
ajax.send(postStr);

//获取执行状态
ajax.onreadystatechange = function() {
//如果执行状态成功，那么就把返回信息写到指定的层里
if (ajax.readyState == 4 && ajax.status == 200) {
msg.innerHTML = ajax.responseText;
}
}
alert (userName);
}
</script>
<body >
<div id="msg"></div>
<form name="user_info" method="post" action="">
姓名：<input type="text" id="user_name"name="user_name" /><br />
年龄：<input type="text" name="user_age" /><br />
性别：<input type="text" name="user_sex" /><br />

</body>

ajax_output.php:

<?php

$username = $_POST['user_name'];
$userage = $_POST['user_age'];
$usersex = $_POST['user_sex'];
echo "$username <br>";
echo "$userage <br>";
echo "$usersex <br>";

$db = new mysqli('localhost','root','123456','test');
if(!$db){
echo "连接失败！";
}
$db->query("set names utf8");
$query = "insert into userinfo(uname,uage,usex) values ('".$username."','".$userage."','".$usersex."')";
$result = $db->query($query);
if ($result){
echo "上传成功！";
}
else {
echo "失败！";
}
$db->close();

?>

第七城市