$str='0501201500';
$date=join('/',substr($str,2,2),substr($str,4,2),20 . substr($str,0,2));
$time=join(':',substr($str,6,2),substr($str,8,2));
$datetime=join(',',$date,$time);
Regards Mark
----- Original Message -----
From: "Jason Balicki" <[EMAIL PROTECTED]>
To: "'Perl Beginners List'" <beginners@perl.org>
Sent: Friday, January 21, 2005 1:04 AM
Subject: More elegant solution for a date conversion
> Hi,
>
> I've got a phone record that keeps the date and time
> in the following format:
>
> YYMMDDHHMM
> example: 0501201500
>
> So, I've written the following to convert it to the
> format:
>
> MM/DD/YYYY,HH:MM
> example: 01/20/2005,15:00
>
>
> sub convertdate {
> my($locdate)[EMAIL PROTECTED];
> $ypfix=20;
> @ardate = split(//,$locdate);
> $year=join('',$ardate[0],$ardate[1]);
> $month=join('',$ardate[2],$ardate[3]);
> $day=join('',$ardate[4],$ardate[5]);
> $hour=join('',$ardate[6],$ardate[7]);
> $minute=join('',$ardate[8],$ardate[9]);
> $jyear=join('',$ypfix,$year);
> $jdate=join('/',$month,$day,$jyear);
> $jtime=join(':',$hour,$minute);
> $retdate=join(',',$jdate,$jtime);
> open ( LOG, ">>$log" );
> print LOG "$retdate";
> close ( LOG );
> }
>
> This works, but strikes me as ugly. Is there a more
> elegant way of doing what I've done here? It seems
> like I should be able to loop through the $ardate[n]
> entries and select them somehow instead of hard coding
> the indexes, but nothing is coming to mind.
>
> TIA,
>
> --J(K)
>
>
> --
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