Re: Formatting elapsed time

[m-rau]

in reply to Formatting elapsed time

The following code is a bit more complex. Sorry. But it produced output reflecting the amount of seconds. If something runs short, it says 3 seconds. If something takes longer, it says 1 hour, 3 seconds. If it takes even longer, the script can say 1 year, 3 monhts, 12 weeks, 4 days, 1 hour, 3 minutes, 20 seconds.
The code is implemented as a runtime tracker.

#!/usr/bin/perl

my $t0;
BEGIN
{
    $t0 = time;
}

END
{
    my $d = time() - $t0;
    my @int = (
        [ 'second', 1                ],
        [ 'minute', 60               ],
        [ 'hour',   60*60            ],
        [ 'day',    60*60*24         ],
        [ 'week',   60*60*24*7       ],
        [ 'month',  60*60*24*30.5    ],
        [ 'year',   60*60*24*30.5*12 ]
    );
    my $i = $#int;
    my @r;
    while ( ($i>=0) && ($d) )
    {
        if ($d / $int[$i] -> [1] >= 1)
        {
            push @r, sprintf "%d %s%s",
                         $d / $int[$i] -> [1],
                         $int[$i]->[0],
                         ( sprintf "%d", $d / $int[$i] -> [1] ) > 1
                             ? 's'
                             : '';
        }
        $d %= $int[$i] -> [1];
        $i--;
    }

    my $runtime = join ", ", @r if @r;
    warn sprintf "RUNTIME %s\n", $runtime;
}

my $runTime = rand( 10 );
printf "Runtime is %d\n", $runTime;
sleep( $runTime );

[5mi11er]

Well, hopefully it takes those 12 weeks and changes that into 2 more months :-)

As a very minor nit, the definitions of year and month may need to be adjusted depending on just how precise one wants to be. There are several definitions/equivalencies for the exact length of a year located here.

For example, using the current Sidereal year's approximation as 365.2564, a month is more accurately said to be 365.2564/12 or 30.438 rather than 30.5, and a year based on 30.5 * 12 is 366 days long...

-Scott