php处理ajax
首先安装wamp,若安装过mysql则终止进程防止冲突,可以访问localhost说明成功。在www目录下新建项目,使用localhost访问。
php:
<?php //3.获取ajax传过来的内容处理 header("content-Type:text/text;charset=utf-8"); $username=$_POST[\'name\']; if($username==\'admin\'){ echo \'{"inf":"该用户名不合法","sta":"0"}\'; } else if($username==\'richard\'){ echo \'{"inf":"该用户名已被注册","sta":"1"}\'; } else{ echo \'{"inf":"该用户名可以注册","sta":"2"}\'; } ?>
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title></title> <style type="text/css"> .green{ color: green; } .red{ color: red; } </style> <script> window.onload=function(){ function id(id){ return document.getElementById(id); } function ajax(){ var oAjax=new XMLHttpRequest(); return oAjax; } id(\'username\').onkeyup=function(){ //1 建立ajax引擎 var xhr=ajax(); //3 php后端操作 var url=\'/m18AjaxT/checkNameJson.php?name=\'+id(\'username\').value; xhr.open(\'POST\',url,true); xhr.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); var data="name="+id(\'username\').value; //2 发送请求,get用null,因为通过url发送 xhr.send(data); xhr.onreadystatechange=function(){ if(xhr.readyState==4 && xhr.status==200){ //4 渲染dom var information=JSON.parse(xhr.responseText).inf; var sta=JSON.parse(xhr.responseText).sta; id(\'inf\').innerHTML=information; if(sta==0){ id(\'inf\').className="red"; }else if(sta==1){ id(\'inf\').className="red"; }else{ id(\'inf\').className="green"; } } } } } </script> </head> <body> <form action="" method="get"> username:<input type="text" id="username" /> <input type="button" name="btn" id="btn" value="验证" /> </form> <span id="inf">this is infomation</span> </body> </html>