Displaying time left to end of day, week, month?

Shell Programming and Scripting
1

Hello,

basically I'm looking after a way of showing how many time is left to the end the day, week, month...

I can't seem to figure this out, so could one of you skilled guys tell me how should this be approached?

Examples:

Time left:

Day: 12h 12m 11s
Week: 4d 12h 12m 11s
Month: 8d 12h 12m 11s

or

Day: 11s
etc....

Thanks for all the help in advance, greetings!

EDIT:

Has to be BASH/SH, as I want to implement this function to one of my scripts where I can extract either time left for day or week or month etc.

2 
#! /usr/bin/perl

use strict;
use warnings;
use integer;
use Time::Local qw{ timelocal_nocheck };

sub timeleft($$@) {
    my $label = shift;
    my $now   = shift;
    my $then  = timelocal_nocheck(@_);

    my $delta_s  = $then - $now;
    my $delta_m  = $delta_s / 60; $delta_s %= 60;
    my $delta_h  = $delta_m / 60; $delta_m %= 60;
    my $delta_d  = $delta_h / 24; $delta_h %= 24;

    local $\ = '';
    local $, = '';

    print $label, ':';

    if (0 < $delta_d) {
        print ' ', $delta_d, 'd';
    }

    if (0 < $delta_d || 0 < $delta_m) {
        print ' ', $delta_h, 'h';
    }

    if (0 < $delta_d || 0 < $delta_m || 0 < $delta_h) {
        print ' ', $delta_m, 'm';
    }

    print ' ', $delta_s, "s\n";
}

my $NOW = time();
my @NOW = localtime($NOW);

$NOW[0] = 0;
$NOW[1] = 0;
$NOW[2] = 0;

my @EOD = @NOW;
   $EOD[3] += 1;

timeleft('  day', $NOW, @EOD);

my @EOW = @NOW;
   $EOW[3] += 7 - $NOW[6]; # wday;

timeleft(' week', $NOW, @EOW);

my @EOM = @NOW;
   $EOM[3]  = 1; # mday
   $EOM[4] += 1; # month

timeleft('month',  $NOW, @EOM);

---------- Post updated at 10:02 AM ---------- Previous update was at 10:01 AM ----------

Example output:

  day: 13h 57m 52s
 week: 5d 13h 57m 52s
month: 9d 13h 57m 52s
3

Thank you m.d.ludwig but I'd need this to be within my bash/sh script, not perl.. Forgot to mention that in the first post, sorry!!!!!

Edit:

Or well preferable in bash but would nobody else step up with an alternative then I can always use your solution too! So again, thanks a lot!

4

Try this...

#!/bin/bash

date_arr=( $(date '+%H %M %S %u %d') )

end_hour=$(echo "23-${date_arr[0]}"| bc)
end_min=$(echo "59-${date_arr[1]}"| bc)
end_sec=$(echo "60-${date_arr[2]}"| bc)

week_end=$(echo "7-${date_arr[3]}"| bc)

month_max=$(cal | awk '/^[0-9]/{d=$NF}END{print d}' )
month_end=$(echo "${month_max}-${date_arr[4]}"| bc)

eod="${end_hour}h ${end_min}m ${end_sec}s"

echo "Day       : $eod"
echo "Week      : ${week_end}d $eod"
echo "Month     : ${month_end}d $eod"

Output

root@bt:/tmp# date
Tue Nov 22 09:27:23 IST 2011
root@bt:/tmp# ./run
Day    : 14h 32m 37s
Week     : 5d 14h 32m 37s
Month     : 8d 14h 32m 37s

HTH
--ahamed

5

Very nice!