sathya83aa has asked for the wisdom of the Perl Monks concerning the following question:

Hi all..

I have a variable in mmddyy format. But I would like to convert it to mm/dd/yyyy format.

Please help me in this regard.

Thanks & Regards,

Sathya V.

Replies are listed 'Best First'.
Re: date conversion
by Rahul6990 (Beadle) on Jul 18, 2013 at 06:34 UTC
    Hi ,

    One option is to use regex:
    my $in_date = '051407';
(my $db_date = $in_date) =~ s/(\d\d)(\d\d)(\d\d)/$1-$2-$3/;
print ":$db_date:";

    [download]

    But there is one issue how would you find that date belongs to 20th century or 21st
    Example:
    The above date could be :05-14-2007 or 05-14-1907

    And one advise its always better to search before you post a question.
[reply]
[d/l]
[reply]
Re: date conversion
by hdb (Monsignor) on Jul 18, 2013 at 07:07 UTC

    You could use substr to insert the slashes and the century into your data as it is already in the correct basic format. For example, substr $data, 2, 0, "/"; would insert a slash after the month into your date. Whether to insert "/20" or "/19" before the year still needs to be decided based on the last two digits of your date (or any other assumption you are willing to make).

[reply]
[d/l]
[select]
Re: date conversion
by Loops (Curate) on Jul 18, 2013 at 07:14 UTC 
    use Date::Manip;
my $in = '051093';
print UnixDate($in =~ s#(..)(..)#$1/$2/#r, '%m/%e/%Y');

    [download]
[reply]
[d/l]
Re: date conversion
by mtmcc (Hermit) on Jul 18, 2013 at 08:25 UTC
    How are you getting your date originally (mmddyy)?

    -Michael

[reply]
Re: date conversion
by Anonymous Monk on Jul 18, 2013 at 07:06 UTC
[reply]

Back to Seekers of Perl Wisdom