[RESOLVED] Resource 23 error

[tommyrulez]

I was getting not a valid mysql function on mysql_fetch_array. Maybe, I am too tired and just can't see it.

<?php
    echo "<table width=\"100%\" border=\"1\">
<tr>   

    <td align=\"center\">Highest rated</td>
    <td align=\"center\">Number of votes</td>
  </tr>";
  include 'db2.php';
  $con;
   $db;
    $mydb;
  $sql = mysql_query("SELECT * FROM results") or die(mysql_error());
$result=mysql_query($sql, $con) or die(mysql_error());
$i=0;
while($row= mysql_fetch_array($result, MYSQL_BOTH)){

   if ($i < mysql_num_fields($result))  {
   echo"<tr>";
  echo "<td align=\"left\">'. $row($i).'</td>";
     echo "<td align=\"left\">'. $row($i).'</td>";
     $i++;
  echo "</tr>";
  }
 } 
echo "</table>";
?>

I can echo $db;, but I can not echo mysql_num_fields($result). So the issue has to be:

$sql = mysql_query("SELECT * FROM results") or die(mysql_error());
$result=mysql_query($sql, $con) or die(mysql_error());

[dagon]

your calling mysql_query twice

$sql = "SELECT * FROM results";
$result=mysql_query($sql, $con) or die(mysql_error()); 

[tommyrulez]

Your suggestion did not work.

[tommyrulez]

update

$sql = "SELECT * FROM 'results'"; 

produces 'query is empty' which is not accurate.

[tommyrulez]

Ok, I removed the and put it back in. Now it works. Sigh*