ajax php用户无刷新登录实例_php教程

<!doctype html public "-//w3c//dtd xhtml 1.0 transitional//en" "http://www.w3.org/tr/xhtml1/dtd/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="content-type" content="text/html; charset=gb2312" />

<title>ajax php用户无刷新登录实例</title>

<script>

function userlogin(){

    var xmlhttp;

    var str;

    var sendstr="";

     try{

             xmlhttp=new xmlhttprequest();

         }

catch(e){

             xmlhttp=new activexobject("microsoft.xmlhttp");

        }

xmlhttp.onreadystatechange=function(){

          if (xmlhttp.readystate==4){

              if (xmlhttp.status==200){

      str = xmlhttp.responsetext;

      document.getelementbyid("userlogin").innerhtml=str;

      }else{

      alert("系统错误,如有疑问,请与管理员联系!"+xmlhttp.status);

    }

          }

       }

xmlhttp.open("post","config/userlogin.php",true);

xmlhttp.setrequestheader('content-type','application/x-www-form-urlencoded');

xmlhttp.send(sendstr);

}

</script>

</head>

<body>

<form id="form1" name="form1" method="post" action="">

  <p>

    <label for="textfield"></label>

    <input type="text" name="uname" id="uname" /><span id="userlogin"></span><br />

<input type="text" name="upwd" id="upwd" /><span id="upwds"></span>

  输入用户名</p>

  <p>

    <input type="button" name="button" id="button" value="登录" onclick="userlogin();" />

  </p>

</form>

</body>

</html>

userlogin.php文件

<?

$uid = $_post['uname'];

$pwd = $_post['upwd'];

$sql ="select * from tabname where uid='$uid' and pwd='$pwd'";

$query = mysql_query( $sql );

if( mysql_num_rows( $query ) )

{

    echo '登录成功';

}

else

{

     echo '用户名或密码不正确!';

  }

?>

(责任编辑：admin)

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ajax php用户无刷新登录实例