[SOLVED] Help with error :Resource id #10

[Darkmatter5]

Here's the code

 

<?php
  if(isset($_POST['code']) && isset($_POST['code'])) {
    $conn=mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to mysql');
      mysql_select_db($dbnameprim);
      $pagedb="members";
                  
      $code=$_POST['code'];
      $cntquery=mysql_query("SELECT COUNT(mem_username)
        FROM $dbnameprim.$pagedb
        WHERE activation_code='$code'
        ") or die(mysql_error());
      list($count)=mysql_fetch_array($cntquery);
      if($count==1) {
        mysql_query("UPDATE $dbnameprim.$pagedb
          SET activated='1'
          WHERE activation_code='$code'
          ") or die(mysql_error());
        $actquery=mysql_query("SELECT mem_username
          FROM $dbnameprim.$pagedb
          WHERE activation_code='$code'
          ") or die(mysql_error());
        $result=mysql_query($actquery) or die($actquery. '<br>' .mysql_error());
        $row=mysql_fetch_array($result);
        //echo "UPDATE $dbnameprim.$pagedb SET activated='1' WHERE activation_code='$code'<br>";
        echo "Thank you for activating your account $row[mem_username]!<br>
        You may now login with your username and password.";
      }
      else {
        echo "That activation code doesn't match any record on file!  Make sure you typed in the correct code.<br>
        If all else fails copy and paste the code from your activation email.";
      }
                  
      mysql_close($conn);
    }              
?>

 

When the ACTIVATE button is pushed it produces the following error.

 

"Resource id #10

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #10' at line 1"

 

Can someone help me troubleshoot this?

[wildteen88]

There error is coming from here, I have highlighted the problem errors

$actquery=mysql_query("SELECT mem_username

          FROM $dbnameprim.$pagedb

          WHERE activation_code='$code'

          ") or die(mysql_error());

        $result=mysql_query($actquery) or die($actquery. '<br>' .mysql_error());

[Twister1004]

When you use mysql_query(), it really returns a reference to a result resource, which is parsed by mysql_fetch_* functions.  If you directly echo out the resource, PHP just gives you the resource number.  You need to actually use one of those mysql_fetch_* functions.