Ajax以post方式提交数据
发布时间:2020-12-16 00:22:44 所属栏目:百科 来源:网络整理
导读:JS script type="text/javascript" function login() { var username = document.getElementById("txtUsername").value; var password = document.getElementById("txtPassword").value; var url = "Login.aspx?op=1"; var postStr = "username="+ username
JS <script type="text/javascript"> function login() {var username = document.getElementById("txtUsername").value; var password = document.getElementById("txtPassword").value; var url = "Login.aspx?op=1"; var postStr = "username="+ username +"&password="+ password; var ajax = null; if (window.XMLHttpRequest) { ajax = new XMLHttpRequest(); } else if (window.ActiveXObject) { ajax = new ActiveXObject("Microsoft.XMLHTTP"); } else { return; } ajax.open("POST",url,true); ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); ajax.send(postStr); ajax.onreadystatechange = function() { if (ajax.readyState == 4 && ajax.status == 200) { document.getElementById("content").innerHTML = ajax.responseText; ; } } } </script> html: <div> 用户名: <input type="text" id="txtUsername" /> 密码: <input type="password" id="txtPassword" /> <input type="button" id="btnOk" value="确定" onclick="login()"/> <div id="content"></div> </div> (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! |