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Home/ Questions/Q 593739
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T15:52:40+00:00 2026-05-13T15:52:40+00:00

I would like to take a time stamp (e.g. 1263531246) and convert it to

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I would like to take a time stamp (e.g. 1263531246) and convert it to a string representation suitable for output to an XML file in an attribute field conforming to xs:dateTime. xs:dateTime expects something like:

2002-05-30T09:30:10-06:00

Ideally, I would use the form of output that includes offset from UTC (as above). In this project, I am constrained to use Perl. Any suggestions?

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  1. Editorial Team
    Editorial Team
    2026-05-13T15:52:40+00:00Added an answer on May 13, 2026 at 3:52 pm

    This works on Linux:

    $ perl -MPOSIX -e 'print POSIX::strftime("%Y-%m-%dT%H:%M:%S%z\n", localtime)'
    2010-02-04T17:37:43-0500

    On Windows, with ActiveState Perl, it prints:

    2010-02-04T17:39:24Eastern Standard Time

    Using DateTime:

    #!/usr/bin/perl
    use strict; use warnings;
    
    use DateTime;
    my $dt = DateTime->now(time_zone => 'EST');
    print $dt->strftime('%Y-%m-%dT%H:%M:%S%z'), "\n"
    

    I get the correct string on Windows as well:

    E:\> t
    2010-02-04T18:06:24-0500

    I believe Date::Format is much lighter weight module:

    #!/usr/bin/perl
    use strict; use warnings;
    
    use Date::Format;
    print time2str('%Y-%m-%dT%H:%M:%S%z', time, 'EST'), "\n";
    

    Output:

    E:\> t
    2010-02-04T18:11:36-0500
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