While the proposed solution below may very well indeed work for this
situation it's a far better practice to strip the variable down to "known
to be good" values rather than "known to be bad" ones. Rather than strip
$ and , marks from the variable it's far better to strip out anything
other than
bsen [mailto:[EMAIL PROTECTED]]
> Sent: Friday, September 14, 2001 7:24 AM
> To: Rick Eicher II
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] currency out of postgresql
>
>
> > $formatted = sprintf("%01.2f", $money);
> > printf ("%01.2f\n", $
AM
To: Rick Eicher II
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] currency out of postgresql
> $formatted = sprintf("%01.2f", $money);
> printf ("%01.2f\n", $formatted);
maybe twice, try:
$formatted = sprintf("%01.2f", $money);
echo "$formatted\n&quo
: <[EMAIL PROTECTED]>
Sent: Friday, September 14, 2001 6:48 PM
Subject: [PHP-DB] currency out of postgresql
> I am trying to get two currency values from out of the database, add them
> together and print the to the screen. I am using the following code.
>
> $money = $myr
I am trying to get two currency values from out of the database, add them
together and print the to the screen. I am using the following code.
$money = $myrow[6] + $myrow[7];
$formatted = sprintf("%01.2f", $money);
printf ("%01.2f\n", $formatted);
$myrow[] is the array that I load the database r
