How do i find-defin-year? How can I become a string and temperature? How can I take seconds? How can I Find the julian day? Does Perl Have A Year 2000 PROBLEM?
How do I find the week-of-a-year / day-of-a-year?
The day of the year is in the array returned by localtime () (see localtime):
$ day_of_year = (localtime (time ())) [7];
Or more legibly (in 5.004 or higher):
Use time :: localtime;
$ day_of_year = localtime (time ()) -> yday;
You can find the week of the year by dividing this by 7:
$ week_of_year = int ($ day_of_year / 7);
Of Course, This Believes That Weeks Start At Zero.
How can I Compare Two Date Strings?
Use the date :: manip or date :: Datecalc Modules from cpan.
How Can I Take A STRING AND TURN IT INTO EPOCH Seconds?
If it's a regular enough string that it always has the same format, you can split it up and pass the parts to timelocal in the standard Time :: Local module. Otherwise, you should look into one of the Date modules from CPAN.
How can I find the julian day?
Neither Date :: Manip nor Date :: DateCalc deal with Julian days. Instead, there is an example of Julian date calculation in http://www.perl.com/CPAN/authors/David_Muir_Sharnoff/modules/Time/JulianDay.pm. GZ, Which Should Help.
Does Perl Have A Year 2000 PROBLEM?
Not unless you use Perl to create one. The date and time functions supplied with perl (gmtime and localtime) supply adequate information to determine the year well beyond 2000 (2038 is when trouble strikes). The year returned by these functions when used in an Array Context is the year minus 1900. for Years Between 1910 and 1999 this happens to be a 2-digit decimal number. to avoid the year 2000 problem simply do not treat the year as a 2-digit number. it isn't.when gmtime () and localtime () are used in a scalar context they return a timestamp string that contains a fully-expanded year. For example, $ timestamp = gmtime (1005613200) sets $ timestamp to `` Tue Nov 13 01:00:00 2001 ''. There's no year 2000 problem.
