Question

求救PHP语法问题，实在是找不到错误在哪里！

<?php
$sql=mysql_query("select * from tb_product where id=".$_GET['id']."",$conn);
$info=mysql_fetch_object($sql);
?>
<div align="center">
<?php
if( $info->img==""){
echo "暂无图片";
}
else
{
?>
<a href="<?php echo $info->img;?>" target="_blank"><img src="<?php echo $info->img;?>" alt="查看大图" width="80" height="80" border="0"></a>
<?php
}
?>
</div>
<div align="center">商品名称：</div>
<div align="left"> <?php echo "$info->mingcheng"; ?></div>

在这段代码中，浏览器提示Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource inE:\EasyPHP-DevServer-14.1VC9\data\localweb\bysj\lookinfo.php on line 3

Notice: Trying to get property of non-object in E:\EasyPHP-DevServer-14.1VC9\data\localweb\bysj\lookinfo.php on line 7
暂无图片
商品名称：

Notice: Trying to get property of non-object in E:\EasyPHP-DevServer-14.1VC9\data\localweb\bysj\lookinfo.php on line 20

请大家帮帮忙谢谢啦！

Answer 1

第二行改成$sql=mysql_query("select * from tb_product where id=".$_GET['id'], $conn);试试