工学结合20191031

mac2024-05-09  7

<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <title>AJAX初识</title> <script type="text/javascript"> window.onload = function() { var btn = document.getElementById("btn"); btn.onclick = function() { //获取用户的用户名和密码 // var uname = document.getElementById("username").value; var pass = document.getElementById("pass").value; //使用ajax发送请求需要如下几步 //1.创建XMLHttpRequest对象 ajax的核心对象 var xhr = new XMLHttpRequest(); //2.准备发送 xhr.open('get','01check.php?username=' + uname + '&password=' + pass, true); //3.执行发送 xhr.send(null); //4.指定回调函数 xhr.onreadystatechange = function() { if (xhr.readyState == 4) { // 4XX 找不到 2XX成功 5XX 服务器错误 if(xhr.status = 200) { var data = xhr.responseText; var info = document.getElementById("info"); if (data == "1") { info.innerHTML = '登录成功' } else { info.innerHTML = '用户名或密码错误'; } } } } } } </script> </head> <body> <form> 用户名: <input type="text" name="username" id="username"> <span id="info"></span> <br /> 密码: <input type="password" name="pass" id="pass"> <input type="button" value="登录" id="btn"> </form> </body> </html>

01check.php↓↓↓

<?php $uname = $_GET['username']; $pass = $_GET['password']; if($uname == "admin" && $padd = "123") { echo "1"; }else{ echo "2"; } ?>
最新回复(0)