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Top Forums Shell Programming and Scripting Substract date (month) Problem Post 302285982 by sunpraveen on Tuesday 10th of February 2009 04:31:50 AM
Old 02-10-2009
Kunal,

Check this link. You might find your answer. Smilie

Request you to go through the FAQ section and search for similar threads before creating a new one.

Regards,

Praveen
 

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GREGORIANTOJD(3)							 1							  GREGORIANTOJD(3)

gregoriantojd - Converts a Gregorian date to Julian Day Count

SYNOPSIS
int gregoriantojd (int $month, int $day, int $year) DESCRIPTION
Valid Range for Gregorian Calendar 4714 B.C. to 9999 A.D. Although this function can handle dates all the way back to 4714 B.C., such use may not be meaningful. The Gregorian calendar was not instituted until October 15, 1582 (or October 5, 1582 in the Julian calendar). Some countries did not accept it until much later. For exam- ple, Britain converted in 1752, The USSR in 1918 and Greece in 1923. Most European countries used the Julian calendar prior to the Grego- rian. PARAMETERS
o $month - The month as a number from 1 (for January) to 12 (for December) o $day - The day as a number from 1 to 31 o $year - The year as a number between -4714 and 9999 RETURN VALUES
The julian day for the given gregorian date as an integer. EXAMPLES
Example #1 Calendar functions <?php $jd = GregorianToJD(10, 11, 1970); echo "$jd "; $gregorian = JDToGregorian($jd); echo "$gregorian "; ?> SEE ALSO
jdtogregorian(3), cal_to_jd(3). PHP Documentation Group GREGORIANTOJD(3)
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