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
$dbquerymeal = select COUNT(*) from mealtype; $resultmeal = mysql_db_query($dbname,$dbqueryshipping1); if(mysql_error()!=){echo mysql_error();} $mealcount = mysql_fetch_row($resultmeal); echo $mealcount; YOUR query is not stored in ,$dbqueryshipping1 but in $dbquerymeal I believe... switch the