RE: [PHP-WIN] Help with Calendar Functions!!

WOW!!! This is the information I needed!!!! I really appreciate your knowledge!!!!! and Help!!!!!! -----Original Message----- From: Zak Greant [mailto:zak@php.net] Sent: Friday, November 10, 2000 10:42 AM To: Joey Garcia Cc: php-windows@lists.php.net Subject: RE: [PHP-WIN] Help with Calendar Functions!! At 10:23 AM 11/10/00 -0700, you wrote: >I figured it was a BC date, but why that date? What is the coorelation >between -4713 and 2000? Would 2001 be -4712? What is the algorithm to >convert it or how do I use -4713, or whatever BC date, to get the current >year? What is funny is that how could it come back with a result of 10/28 >(month/day) for Julian Day of 301 when it doesn't know the year? (Leap >year can throw that off) Julian Day Counts are not what you think they are. :) A JD count is the number of days that have elapsed since November 25th, 4714 BC >I tried your example but that just gave me a result of "2451859", but no >matter what I do I can't get that to convert to the current year(2000). >Again, I don't know the correct algorithm, I get either 1999 or 2004 >depending if I use 365 or 365.25. > >Can you get a Julian day from a date using these functions? > >Example, > >For Date: 1/1/2000 the Julian Day: 1 >For Date: 3/1/200 the Julian Day: 61 >etc. ... To get the number of days cound for the year using JD counts, do something like this: list ($m, $d, $y) = explode ('/', date ('m/d/Y')); $new_years_day = gregoriantojd (1, 1, $y); $today = gregoriantojd ($m, $d, $y); print ($today - $new_years_day) + 1; Or just use date ('z') :) >Again, I really appreciate your help!!! No prob - I have been doing a lot of work with the calendar functions lately. :) --zak