Php 警告:mysql_fetch_object():提供的参数在json中不是有效的mysql结果资源
我的json代码中有一个错误,如下所示 错误: 警告:mysql\u fetch\u 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”:null} PHP文件:Php 警告:mysql_fetch_object():提供的参数在json中不是有效的mysql结果资源,php,mysql,json,Php,Mysql,Json,我的json代码中有一个错误,如下所示 错误: 警告:mysql\u fetch\u 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”:null} PHP文件: <?php @include("db.php"); $query = "SELECT a.a_name as name,b.total_value as value,b.total_votes as vot
<?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);
?>
MY 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($link);
?>
有人能帮我吗?我想问题在下面的部分询问:
inner join ratings b on b.a_id=a.id"
从您的查询中,我猜a_u[some_utext]是表a的列,但是您的查询在表b的a_id上进行连接。我认为应该是这样的:
inner join ratings b on a.a_id=b.id"
您是否尝试了mysql_error()?这样地。。。如果(!$result){die('Invalid query:'.mysql_error());},明显的原因是数据库连接失败或查询失败。从include语句中删除错误化脓符号。若要获得有关sql错误的警告,请尝试此操作,如果(!$result)死亡(“mySQL错误:.mySQL_error()”),则可帮助查找sql错误@shuvo-非常感谢你。。我犯了错误,所以呢?你的错误是什么。。。写在这里,以便对有类似问题的人有所帮助。