时间：2020-12-02 12:29:23 | 栏目：JavaScript代码 | 点击：次
本文实例为大家分享了js实现ajax的用户简单登入的具体代码，供大家参考，具体内容如下
原生js实现ajax
html页面
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>ajax登录</title>
</head>
<body>
<div>
<div id="showInfo"></div>
<form id="form">
用户名：<input type="text" name="username" id="username"><br>
密码：<input type="password" name="password" id="password">
<input type="button" id="btn" value="登录">
</form>
</div>
<script type="text/javascript">
window.onload = function(){
var btn = document.getElementById('btn');
btn.onclick = function(){
var username = document.getElementById('username').value;
var password = document.getElementById('password').value;
//第一步：创建对象
var xhr = null;
if(window.XMLHttpRequest){
xhr = new XMLHttpRequest();
}else{
xhr = new ActiveXObject("Microsoft.XMLHTTP");
}
//初始化
//准备好了
var url = './check.php?username='+username+"&password="+password;
xhr.open('post',url,false);
//这段代码在xhr.send();执行完之后才能执行
//这件事做完了怎么办
//事情办完之后干什么
xhr.onreadystatechange = function(){
if(xhr.readyState == 4){
if(xhr.status == 200){
alert(1);
var data = xhr.responseText;
if(data == 1){
document.getElementById('showInfo').innerHTML = '用户名或者密码错误';
}else if(data == 2){
document.getElementById('showInfo').innerHTML = '登录成功';
}
}
};
}
//实际的去做这件事
//去做这件事情
xhr.send(null);
alert(2);
}
}
</script>
</body>
</html>
check.php
<?php
$username = $_GET['username'];
$password = $_GET['password'];
if($username == 'admin' && $password == '123'){
echo 2;
}else{
echo 1;
}
?>