Hi all, I have wriiten a function that determines whether tomorrows date is the first of the month or not then returns a SQL string based on whether its the first of the month or not. According to my apache error logs I get an error that says:
[Sun Mar 26 21:43:14 2006] [error] [client 192.168.0.2] PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\\Program Files\\Apache Group\\Apache2\\htdocs\\validation.php on line 331 I understand that this is a warning but I believe that it has something to do with the fact that no output is being displayed. All other mysql database outputs work fine. This is the code that sets the query in the mysql_query parameter $Month_query = mysql_query($this->determineMonth()); this is the code for determineMonth () function determineMonth() { $return = ""; $query1 = mysql_query("SELECT months FROM Month WHERE m_id = month(curdate())"); $query2 = mysql_query("SELECT dayNum FROM Days WHERE dayNum = day(curdate())"); $query3 = mysql_query("SELECT year FROM Year WHERE year = year(curdate())"); switch ($query1) { case "January": case "March": case "May": case "July": case "August": case "October": case "December": if($query2 == 31) $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())+1"; else $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())"; break; case "February": if ($query2 == 28 || $query2 == 29) $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())+1"; else $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())"; break; case "April": case "June": case "September": case "November": if ($query2 == 30) $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())+1"; else $return = "SELECT m_id, months FROM Month WHERE m_id >= month(curdate())"; break; } return $return; } I hope I included everything. If not let me know and I'll post it Paul -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php