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 =
! 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 =
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 =
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
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
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