php處理ajax
阿新 • • 發佈：2017-05-18
.class request 通過 action get check parse ready inf
首先安裝wamp,若安裝過mysql則終止進程防止沖突，可以訪問localhost說明成功。在www目錄下新建項目，使用localhost訪問。
php:
<?php //3.獲取ajax傳過來的內容處理 header("content-Type:text/text;charset=utf-8"); $username=$_POST[‘name‘]; if($username==‘admin‘){ echo ‘{"inf":"該用戶名不合法","sta":"0"}‘; } else if($username==‘malin‘){echo ‘{"inf":"該用戶名已被註冊","sta":"1"}‘; } else{ echo ‘{"inf":"該用戶名可以註冊","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後端操作 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 發送請求，get用null,因為通過url發送 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="驗證" /> </form> <span id="inf">this is infomation</span> </body> </html>
php處理ajax