警告:mysql_fetch_object(）:提供的参数不是 json 中的有效 MySQL 结果资源

我的 JSON 代码中出现了这样的错误。

错误：

警告：mysql_fetch_object（）：提供的参数不是第 15 行 D：''XAMPP''xampp''htdocs''ROOPA''music''demo.php 中的有效 MySQL 结果资源{"结果"：空}

PHP文件：

<?php
 @include("db.php");

  $query = "SELECT a.a_name as name,b.total_value as value,b.total_votes as votes,a.a_pic as image FROM _album a inner join ratings b on b.a_id=a.id"; 
  $result = mysql_query($query);
//  $query1 = "SELECT total_value,total_votes FROM ratings";
  //$result1 = mysql_query($query1);
  $count = mysql_num_rows($result);
    //$count1 = mysql_num_rows($result1);
   if($count > 0)
    {
        while($data = mysql_fetch_object($result))
         {
              $alb_name =$data->name; 
              $rate_value = $data->value;
              $rate_votes = $data->votes;
              $alb_pic =$data->image; 
              $resmsg[] = array("Album_name"=>$alb_name,"Rating_total_value"=>$rate_value,"Rating_total_votes"=>$rate_votes,"Image_name"=>$alb_pic);
         }
        $jsonarr = array("result"=>$resmsg);
    }
    else
    {
        $jsonarr = array("result"=>"data not found");
    }
 echo json_encode($jsonarr);
?>  

我的数据库。PHP文件：

<?php
$hostname="localhost";
$username="root";
$password="";
$database="musicalbum";
$conn=mysql_connect($hostname,$username,$password,$database);
$link=mysql_select_db($database,$conn);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
@mysql_close($link);
?>

谁能帮我？