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