I would be using Time::Piece for this since it is in core.

#!/usr/bin/env perl
use strict;
use warnings;
use Time::Piece;

my $t = Time::Piece->strptime ($ARGV[0], '%d/%m/%Y');
printf "Date as supplied: %s\n", $t->dmy ('/');
$t = $t - 86400;
printf "Date minus one day: %s\n", $t->dmy ('/');
[download]

$ ./yesterday.pl 01/12/2022
Date as supplied: 01/12/2022
Date minus one day: 30/11/2022
$
[download]

Note that this approach will have a problem for those days with less than 24 hours (like when switching between normal and daylight savings time.

I prefer the approach to iteratively subtract 22 hours until the (stringified) date changes:

...

my $res = $t - 22*60*60;
while( $res->ymd('-') eq $t->ymd('-') ) {
    $res -= 22*60*60;
}
return $res
[download]

Note that this approach will have a problem for those days with less than 24 hours

I would be interested to see a test case showing that, if you could provide one. The tests I tried all worked fine, even on 23-hour days. That's why I posted it as-is rather than trying to work around a problem which didn't appear to be there.

Here's my trivial test script for comparison:

use strict;
use warnings;

use Time::Piece;
use Test::More tests => 366 * 2 + 1;

my $dstr = '01/01/2023';

for (1 .. 366) {
    my $t = Time::Piece->strptime ($dstr, '%d/%m/%Y');
    my $nt = $t - 86400;
    isnt $dstr, $nt->dmy ('/');
    my $ddiff = $t->mday - $nt->mday;
    ok ($ddiff < 0 || $ddiff == 1), '1 day different or month wrap';
    $dstr = $nt->dmy ('/');
}
is $dstr, '31/12/2021';
[download]

---

🦛

> Note that this approach will have a problem for those days with less than 24 hours

If you want Time::Piece to respect time zones, you need to initialize the object from localtime, not from the class itself:

$ TZ=Europe/London perl -MTime::Piece -wE '
    for my $source ("Time::Piece", scalar localtime()) {
        my $t = $source->strptime("2023/03/27", "%Y/%m/%d");
        say $t - 86400;
    }'
Sun Mar 26 00:00:00 2023
Sat Mar 25 23:00:00 2023
[download]

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

Note that this approach will have a problem for those days with less than 24 hours

That's not true. Or rather, there are no days with less than 24 hours when dealing with UTC, as the parent code does.

G'day hippo,

++ I agree with your approach; particularly in light of the OP's "I am not concerned about daylight savings time, time zone nor do I need HMS.".

Having said that, I believe use of the ONE_DAY constant from Time::Seconds (also a core module) would add clarity.

It's the same value:

$ perl -E 'use Time::Seconds; say ONE_DAY'
86400
[download]

It works as a direct replacement in your code:

$ perl -e '
    use strict;
    use warnings;

    use Time::Piece;
    use Time::Seconds;

    my $t = Time::Piece->strptime($ARGV[0], "%d/%m/%Y");
    printf "Date as supplied: %s\n", $t->dmy("/");
    $t = $t - ONE_DAY;
    printf "Date minus one day: %s\n", $t->dmy("/");
' 01/12/2022
Date as supplied: 01/12/2022
Date minus one day: 30/11/2022
[download]

— Ken

If $Time::Seconds::ONE_DAY remains constant regardless of current time, then do not see how that would add "clarity" or see its purpose. If 24 *60 *60 conjures up magic number, then just assign that to a local variable.

With DateTime (DateTime::Format::Strptime):

use strict;
use warnings;
use DateTime::Format::Strptime;

my $parser = DateTime::Format::Strptime->new(pattern => '%d/%m/%Y', on
+_error => 'croak');

my $dt = $parser->parse_datetime('28/05/2023');

print $dt->subtract(days => 1)->strftime('%d/%m/%Y'), "\n";
[download]

Hope this helps!

Avoid all those problems with short or long days and daylight savings time switches by using the middle of the day :)

#!/usr/bin/perl

use strict; # https://perlmonks.org/?node_id=11152433
use warnings;
use Time::Piece;
use Time::Seconds;

@ARGV = '28/05/2023';

my $t = -ONE_DAY + Time::Piece->strptime("$ARGV[0] 12", "%d/%m/%Y %H")
+;
print localtime($t) . " should be near middle of the day\n\n";
printf "Date as supplied: %s\nDate minus one day: %s\n", $ARGV[0], $t-
+>dmy("/");
[download]

Outputs:

Sat May 27 12:00:00 2023 should be near middle of the day

Date as supplied: 28/05/2023
Date minus one day: 27/05/2023
[download]

Confusing: 04/24/2023 and 05/31/2023 are a direct contradiction to the DD/MM/YYYY given at start of post. Which is it?

Hello, what I meant was it can be any date that the script will provide - I was just giving examples. The script format will display MMDDYYYY. From there I need to know how to figure out what the day before was. As to my examples the day before would be 04/23/2023 and 05/30/2023 respectively.