php - 警告 : mysql_fetch_object(): supplied argument is not a valid MySQL result resource in json

Question

我的 json 代码中有一个错误，如下所示。

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 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);

?>

有人可以帮助我吗？

Answer 1

我认为问题在查询的以下部分:

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

根据您的查询，我猜测 a_[some_text] 是表 a 的列，但您的查询连接表 b 中的 a_id 。我想，应该是这样的:

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