[Solved] Subtracting no of days with system date and converting to MMDDYYYY format in perl [closed]

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I was close to closing this question as you didn’t include your workings. But I’ve seen your comment – you should edit your question to include that.

So here’s your code:

use POSIX 'strftime';
print strftime "%Y_%m_%d_%H_%M_%S/n", localtime() - 24 * 60 * 60);

That’s pretty close. You have three small issues.

  1. You have /n where you meant \n.
  2. You have 24 * 60 * 60, which is one day. You wanted 45 *24 * 60 * 60.
  3. You need to pass subtract the 45 days from the output from time() and then pass that value to localtime().

So using your method, you would want:

use POSIX 'strftime';
print strftime "%Y_%m_%d_%H_%M_%S\n", localtime(time - 45 * 24 * 60 * 60;

But I’d write it using Time::Piece and Time::Seconds (both of which have been part of the Perl code since 5.10.

use 5.010; # for say()
use Time::Seconds;
use Time::Piece

say localtime(time - 45 * ONE_DAY)->strftime("%Y_%m_%d_%H_%M_%S");'

Update: If you’re going to be running this in the hours around a DST changeover, then you’re going to get the wrong answer. The usual approach to fixing this is to normalise the localtime to noon before doing the subtraction.

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solved Subtracting no of days with system date and converting to MMDDYYYY format in perl [closed]