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

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(mysq

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"; > $resultmeal = mysql_db_query($dbname,$dbquer

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 = mysql_fe