[PHP-DB] Code problem - need help.

2003-01-12 Thread Ron Clark
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 =

Re: [PHP-DB] Code problem - need help.

2003-01-12 Thread Chase Urich
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? The mysql_query command returns a result set, not the