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Reg expression on day

by Anonymous Monk
on Dec 23, 2003 at 16:49 UTC ( #316653=perlquestion: print w/replies, xml ) Need Help??

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

I need to substitute a date entry if the day has a "0" as the beginning. For example if I find this entry: 12/09/2003 I need it to be: 12/9/2003 Please advise:
s/^\d{2}\/(0)\d\/\d{4}//g;

Replies are listed 'Best First'.
Re: Reg expression on day
by Thelonius (Priest) on Dec 23, 2003 at 16:59 UTC
    That would delete the date entirely. Here's one that would work:
    s!^(\d+/)0(\d/\d{4})!$1$2!;
      Thank you!
Re: Reg expression on day
by Roy Johnson (Monsignor) on Dec 23, 2003 at 17:19 UTC
    s!/0!/!

    Update:
    To remove all leading zeroes (day, month, and year):
    s/\b0//g


    The PerlMonk tr/// Advocate
Re: Reg expression on day
by flounder99 (Friar) on Dec 23, 2003 at 17:59 UTC
    If you only want to get rid of the leading zero on the day this would work:
    s!/0?(\d+)/!/$1/!
    If you want to get rid of the leading zeros on both the month and the day this would work:
    s!0?(\d+)/0?(\d+)!$1/$2!
    use strict; for ('12/09/2003', '12/10/2003', '01/01/2004') { my $date = $_; print "Before: $date\n"; $date =~ s!/0?(\d+)/!/$1/!; print "Leading zero off of day: $date\n"; $date = $_; $date =~ s!0?(\d+)/0?(\d+)!$1/$2!; print "Leading zero off of month and day: $date\n\n"; } __END__ Before: 12/09/2003 Leading zero off of day: 12/9/2003 Leading zero off of month and day: 12/9/2003 Before: 12/10/2003 Leading zero off of day: 12/10/2003 Leading zero off of month and day: 12/10/2003 Before: 01/01/2004 Leading zero off of day: 01/1/2004 Leading zero off of month and day: 1/1/2004

    --

    flounder

      you can replace
      $date =~ s!0?(\d+)/0?(\d+)!$1/$2!;
      with
      $date =~ s!0?(\d+/)!$1!g;
      for month and day.
Re: Reg expression on day
by Paulster2 (Priest) on Dec 23, 2003 at 17:55 UTC

    An alternate way of doing this might be to break out your MM/DD/YYYY into seperate chunks (ie $MM = month, $DD = day, and $YYYY = year), then take the day variable and add zero to it. Then put it all back together again. This way you could manipulate the date if you needed to.

    ($MM, $DD, $YYYY) = split(/,"$date"); $MM=$MM+0; $DD=$DD+0; $date=$MM."/".$DD."/".$YYYY;

    Please forgive me if my split syntax is incorrect. I don't have an example in front of me. Anyway, I am just providing an alternate methodry. Although if you absolutely need to use a regex or want to get better at using them, this definitely won't help you!

    Have fun!

    UPDATE: Thanks flounder99 for the correction, I thought that I was missing something!

    Paulster2

      Please forgive me if my split syntax is incorrect

      All you need is to quote your slash:

      split('/',"$date")
      I was thinking of something similar:
      s!(\d+)/(\d+)!($1+0).'/'.($2+0)!e

      --

      flounder

      Sure, and it cleans up a bit:

      $date = join "/", map $_+0, split m[/], $date;

      Not sure if I like the idea of stripping those zeroes though. Probably better to store the date in a saner format (Epoch seconds maybe?), then use POSIX::strftime() to generate the desired date format.
Re: Reg expression on day
by jbware (Chaplain) on Dec 23, 2003 at 18:40 UTC
    Here's another way to do it, and it works for leading zeros on month and day.
    s|0(\d)/|$1/|g;
      Or more simply: s!\b0!!g.

      The PerlMonk tr/// Advocate
      Again, thanks for all the input!
Re: Reg expression on day
by Art_XIV (Hermit) on Dec 23, 2003 at 20:57 UTC

    Here's a way that makes gratuitous use of map! ;)

    use strict; while (<DATA>) { chomp; my $val = join '/', map {$_ + 0} (split /\//, $_); print "$val\n"; } __DATA__ 12/09/2004 01/15/2003 03/09/2004 10/28/2002 004/00003/02003

    Isn't Perl wonderful?

    Hanlon's Razor - "Never attribute to malice that which can be adequately explained by stupidity"
Re: Reg expression on day
by delirium (Chaplain) on Dec 24, 2003 at 13:21 UTC
    Since this is already getting out of hand....

    substr($_,3,2) += 0;

    This will work only if the month always has two digits.

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