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))
it's probobly your sql sommand. Have you printed $qry to the screen and ran it straight into the db to see what error it gives. I would bet this is the problem.
Actually I got the querys working. I was just missing a comma in the first one. I also ran both querys through command line and they work fine. My only problem now is that when I open the script through my browser it only displays a blank page. I am a total newbie at this stuff so any help would be greatly appreciated.