WARNING: supplied argument not valid

[DRTechie]

I am getting an error message on a code someone supplied for me and I can't figure what's wrong with it. Maybe a fresh pair of eyes can spot where the error is coming from. Thanks.

Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/circuit_report.php on line 18

// sites that have at least 2 primary type circuits
$sql = "SELECT sc.site_id, COUNT(sc.site_id) AS total_circuits
FROM site s, circuits c, site_circuit sc
WHERE type = 'primary' AND
sc.record_id = c.record_id AND sc.site_id = s.site_id 
GROUP BY sc.site_id
HAVING total_circuits >= 2";

$site_result = mysql_query($sql);

while ($site_array = mysql_fetch_array($site_result))
{
//select details for site_id's returned in first query 
$site_id = $site_array['site_id'];
$sql = "SELECT sc.ckt_id, sc.site_id,  s.sitename, s.address1 
FROM site_circuit sc, circuits c, site s 
WHERE type = 'primary' AND sc.site_id = '$site_id' 
AND sc.record_id = c.record_id 
AND sc.site_id = s.site_id";

$detail_result = mysql_query($sql);
while ($detail_array = mysql_fetch_array($detail_result))
{ 
// get details of each site
// repeated rows for each row in relation
echo $site_detail['sitename'];
}
}

LINE 18 is

while ($site_array = mysql_fetch_array($site_result))

[hand]

Add "or print ..."-statements to get sql error messages:

// sites that have at least 2 primary type circuits
$sql = "SELECT sc.site_id, COUNT(sc.site_id) AS total_circuits
   FROM site s, circuits c, site_circuit sc 
   WHERE type = 'primary' AND
   sc.record_id = c.record_id AND sc.site_id = s.site_id 
   GROUP BY sc.site_id
   HAVING total_circuits >= 2";

$site_result = mysql_query($sql)
   or print "&gt;&gt;&gt; MySQL-Error: ".mysql_errno()." -&gt; ".mysql_error()."<br>\n";

while ($site_array = mysql_fetch_array($site_result)) {
   //select details for site_id's returned in first query 
   $site_id = $site_array['site_id'];
   $sql = "SELECT sc.ckt_id, sc.site_id,  s.sitename, s.address1 
      FROM site_circuit sc, circuits c, site s 
      WHERE type = 'primary' AND sc.site_id = '$site_id' 
      AND sc.record_id = c.record_id 
      AND sc.site_id = s.site_id";

   $detail_result = mysql_query($sql)
      or print "&gt;&gt;&gt; MySQL-Error: ".mysql_errno()." -&gt; ".mysql_error()."<br>\n";

   while ($detail_array = mysql_fetch_array($detail_result)) { 
      // get details of each site
      // repeated rows for each row in relation
      echo $site_detail['sitename'];
   }
}

[DRTechie]

I got it, I was just missing a comma in my first query. The problem now is that when I open the script in my browser it doesn't display anything. I just get a blank page.