09-02-2013
Date format in Bash Script
Hi Experts,
We get "Day" of a month in a variable, so how to make date of out it?
To make more sense
if my variable $DAY contains "12" and month and year will be current date (as of today)
so I want to see as output as 2013-09-12.
How can I achive this bash script??
Any help is highly appreciated.
Thank you
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LEARN ABOUT PHP
cal_from_jd
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)