On 23 Jul 2003 at 15:38, Phillip Blancher wrote: > Problem with Count. > > ! am trying to count the number of items in this table. The table has > ! one field in it. > > The code I am using is: > > $dbquerymeal = "select COUNT(*) from mealtype"; > $resultmeal = mysql_db_query($dbname,$dbqueryshipping1);
shouldn't the second argument be $dbquerymeal ? R'twick -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
