Ajax -post 请求
发布时间:2020-12-16 02:58:56 所属栏目:百科 来源:网络整理
导读:!DOCTYPE htmlhtml lang="en"head meta charset="UTF-8" titleAJAX发送POST请求/title style #loading { display: none; position: fixed; top: 0; left: 0; right: 0; bottom: 0; background-color: #555; opacity: .5; text-align: center; line-height: 3
<!DOCTYPE html><html lang="en"><head> <meta charset="UTF-8"> <title>AJAX发送POST请求</title> <style> #loading { display: none; position: fixed; top: 0; left: 0; right: 0; bottom: 0; background-color: #555; opacity: .5; text-align: center; line-height: 300px; } #loading::after { content: ‘加载中...‘; color : #fff; } </style></head><body><div id="loading"></div><table border="1"> <tr> <td>用户名</td> <td><input type="text" id="username"></td> </tr> <tr> <td>密码</td> <td><input type="password" id="password"></td> </tr> <tr> <td></td> <td><button id="btn">登录</button></td> </tr></table><script> // 找一个合适的时机,做一件合适的事情 var btn = document.getElementById(‘btn‘); // 1. 获取界面上的元素 value var txtUsername = document.getElementById(‘username‘); var txtPassword = document.getElementById(‘password‘); var loading = document.getElementById(‘loading‘); btn.onclick = function () { loading.style.display = ‘block‘; var username = txtUsername.value; var password = txtPassword.value; // 2. 通过 XHR 发送一个 POST 请求 var xhr = new XMLHttpRequest(); xhr.open(‘POST‘,‘login.php‘); // !!! 一定注意 如果请求体是 urlencoded 格式 必须设置这个请求头 !!! xhr.setRequestHeader(‘Content-Type‘,‘application/x-www-form-urlencoded‘); // xhr.send(‘username=‘ + username + ‘&password=‘ + password) xhr.send(`username=${username}&password=${password}`); // 3. 根据服务端的反馈 作出界面提示 xhr.onreadystatechange = function () { if (this.readyState !== 4) return; console.log(this.responseText); loading.style.display = ‘none‘ } }</script></body></html> (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |