Question

我的json代码中出现了错误，。

ERROR：

警告：mysql_fetch_object（）：提供的参数不是第15行的D：\ XAMPP \ xampp \ htdocs \ ROOPA \ music \ demo.php中的有效MySQL结果资源 {“result”：null}

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DB.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);

?>

任何人都可以帮助我吗？

Answer 1

我认为问题的下面部分存在问题：

inner join ratings b on b.a_id=a.id"

从您的查询中，我猜a_ [some_text]是表a的列，但您的查询从表b加入a_id。我想，它应该如下：

inner join ratings b on a.a_id=b.id"

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

1 个答案: