PHP怎么使用AJAX返回登录成功信息！

<code>

    
        <meta charset="utf-8">
<title>登录</title>
<script type="text/javascript" src="http://cdn.bootcss.com/jquery/2.1.4/jquery.min.js"></script>
                        帐号：<input type="text" id="account"><br><br>
                        密码：<input type="password" id="password"><br><input type="button" value="登录" id="btnlogin"><script type="text/javascript">
        $(function(){
            $("#btnlogin").click(function(){
                $.ajax({
                    type:"post",
                    url:"ajax.php",
                    data:{account:$("#account").val(),password:$("#password").val()},
                    dataType:"json",
                    success:function(data){
                        if(data.type==1){
                            alert("登录成功");
                        }else{
                            alert("登录失败");
                        }
                    },
                    error:function(){
                        alert("请求异常");
                    }
                });
            });
        });
        </script></code>

<code><?php header("Content-Type:text/html; charset=utf-8");
$account = $_POST['account'];
$password = $_POST['password'];
$result = array();
if ($account != '' && $password != '') {
    //$row = $db->query("SELECT * FROM account where user = '".$account."' and password = '".$password."'");
    $row = true;//这里去查数据库，假设这里返回true
    if($row){
        $result['type'] = 1;
        $result['msg'] = '登录成功';
    }else{
        $result['type'] = 0;
        $result['msg'] = '用户名或密码不正确';
    }
} else {
    $result['type'] = 0;
    $result['msg'] = '参数传输不正确';
}
echo json_encode($result);
?&gt;</code>

<code>header('content-type:application/json; charset=utf-8');
exit(json_encode($yourdata));</code>