原生js实现ajax的用户简单登入

原生js实现ajax

  • html页面

<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
 
$username = $_GET['username'];
$password = $_GET['password'];

if($username == 'admin' && $password == '123'){
    echo 2;
}else{
    echo 1;
}


?>

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