Question

Well I have been on this problem the entire day and I think that qualifies the need for some help.

PHP

$hour1 = (float) $date1 / 10000;
$hour1 = (float) floor($hour1);

$hour2 = (float) $date2 / 10000;
$hour2 = (float) floor($hour2);

$minute1 = (float) $date1 / 10000;
$minute1 = (float) $minute1 - floor($minute1);
$minute1 = (float) $minute1 / 60;
$minute1 = (float) $minute1 * 100;

$minute2 = (float) $date2 / 10000;
$minute2 = (float) $minute2 - floor($minute2);
$minute2 = (float) $minute2 / 60;
$minute2 = (float) $minute2 * 100;

$date = (float) ($hour1 + minute1) - ($hour2 + minute2);
$date = floatval($date);

As you can see I’ve went a little overboard with the type casting.

Variable Values

$minute1 = .564478
$hour1 = 19

$minute2 = .9885546
$hour2 = 8

$date = (float) ($hour1 + minute1) - ($hour2 + minute2);

IS

$date = (float) (19 + .564478) - (8 + .9885546);

Now the result given is actually

11

AND NOT

10.575932......

Curiously, when I do this without variables echo (19 + .564478) - (8 + .9885546);… I get the floating point result of 10.575932.......

I understand that most programming languages round out floating points but I’ve tried everything including up to explicitly type casting. What is going on here?

Answer 1

$date = (float) ($hour1 + minute1) - ($hour2 + minute2);

Didn’t you forget about $ before minute1 and minute2 ??

If you had, PHP would treat this vars as constans and count only $hour1 - $hour2 witch returns 11 in your case

it should look like

$date = (float) ($hour1 + $minute1) - ($hour2 + $minute2);