! 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
$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
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