Code problem - need help.
| From: | Ron Clark | Date: | Sun, 12 Jan 2003 08:30:35 +0000 |
| Subject: | Code problem - need help. | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-24782@lists.php.net to get a copy of this message | ||
Hello all,
I have the following code:
echo "<td>\n";
echo "<table width=\"100%\" border=\"0\"
align=\"center\"
cellpadding=\"2\">\n";
$sql2 = mysql_query("SELECT signature FROM event ORDER BY cid DESC
LIMIT 10");
while ($row = mysql_fetch_array($sql2)) {
// $cid_alert = $row["cid"];
$number = $row["signature"];
$name = mysql_query("SELECT sig_name FROM signature WHERE sig_id =
$number");
echo
"<tr><td><p>$number</p></td><td><p>$name</p></td></tr>\n";
}
echo "</table>\n";
echo "</td>\n";
When I view on my web page, the $number value comes out fine, but the $name
comes out as "Resource ID #x" where x is some number. Currently is says
"Resource ID #6". I know the SQL syntax is right, I have tested it. But what
is wrong here?
I have tried putting single quotes around $number in my query statement. It
dod not work though. If anyone can help, it is greatly appreciated!!!
Thanks,
R Clark