警告：mysql_query（）期望参数2是资源，第103行给出的字符串[重复]

Hello I AM getting this error i am a newbie pls help..

<?php
$queryyy=mysql_query($conn,"select * from comments where type='".$platform."' AND app_id=".$id." order by id DESC");
$queryyy = mysql_query($sql);
if (!$queryyy) {
   die('Invalid query: ' . mysql_error());
   }


while($rowww=mysql_fetch_array($queryyy))
{
echo $rowww['comm']."<hr/>";
}
?>

Here is my code..

</div>

try this

    <?php
$queryyy=mysql_query("select * from comments where type='".$platform."' AND app_id=".$id." order by id DESC") or die(mysql_error());//it will give you the right error



while($rowww=mysql_fetch_array($queryyy))
{
echo $rowww['comm']."<hr/>";
}
?>

Learn mysqli or PDO.

And in MYSQL connection is optional and its second parameter not first.

Try this ...

<?php
//use your connection file here...
$queryyy=mysql_query("select * from comments where type='$platform' AND app_id='$id' order by id DESC");
if (!$queryyy) {
   die('Invalid query: ' . mysql_error());
   }
while($rowww=mysql_fetch_array($queryyy))
{
echo $rowww['comm']."<hr/>";
}
?>