[PHP] newbY prob

2003-07-23 Thread Phillip Blancher
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); if(mysql_error()!=){echo mysql_error();} $mealcount =

RE: [PHP] newbY prob

2003-07-23 Thread Jennifer Goodie
! 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); if(mysql_error()!=){echo mysql_error();} $mealcount =

Re: [PHP] newbY prob

2003-07-23 Thread R'twick Niceorgaw
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 =

Re: [PHP] newbY prob

2003-07-23 Thread CPT John W. Holmes
From: Jennifer Goodie [EMAIL PROTECTED] ! 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); if(mysql_error()!=){echo

RE: [PHP] newbY prob

2003-07-23 Thread Carl Furst
at your own risk. Carl Furst Chief Technical Officer Vote.com 50 Water St. South Norwalk, CT. 06854 203-854-9912 x.231 -Original Message- From: Phillip Blancher [mailto:[EMAIL PROTECTED] Sent: Wednesday, July 23, 2003 3:38 PM To: PHP List Subject: [PHP] newbY prob Problem with Count. ! am