How to get year part from file created date?

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Old 01-15-2014
Originally Posted by Yoda
Use -E option:
-E           The same as -l, except  displays  time  to  the
             nanosecond  and  with  one format for all files
             regardless  of age: yyyy-mm-dd hh:mm:ss.nnnnnnnnn (ISO 8601:2000 format).

Thanks, this works perfectly
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gregoriantojd - Converts a Gregorian date to Julian Day Count

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)