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Full Discussion: if-statement troubles
Top Forums Shell Programming and Scripting if-statement troubles Post 302341778 by doc.arne on Thursday 6th of August 2009 02:47:12 PM
Old 08-06-2009
Easy enough, but it worked. My book said one = should be enough
 

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CAL_FROM_JD(3)								 1							    CAL_FROM_JD(3)

cal_from_jd - Converts from Julian Day Count to a supported calendar

SYNOPSIS
array cal_from_jd (int $jd, int $calendar) DESCRIPTION
cal_from_jd(3) converts the Julian day given in $jd into a date of the specified $calendar. Supported $calendar values are CAL_GREGORIAN, CAL_JULIAN, CAL_JEWISH and CAL_FRENCH. PARAMETERS
o $jd - Julian day as integer o $calendar - Calendar to convert to RETURN VALUES
Returns an array containing calendar information like month, day, year, day of week, abbreviated and full names of weekday and month and the date in string form "month/day/year". EXAMPLES
Example #1 cal_from_jd(3) example <?php $today = unixtojd(mktime(0, 0, 0, 8, 16, 2003)); print_r(cal_from_jd($today, CAL_GREGORIAN)); ?> The above example will output: Array ( [date] => 8/16/2003 [month] => 8 [day] => 16 [year] => 2003 [dow] => 6 [abbrevdayname] => Sat [dayname] => Saturday [abbrevmonth] => Aug [monthname] => August ) SEE ALSO
cal_to_jd(3), jdtofrench(3), jdtogregorian(3), jdtojewish(3), jdtojulian(3), jdtounix(3). PHP Documentation Group CAL_FROM_JD(3)
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