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Top Forums Shell Programming and Scripting meaning of today=${1:-${today}} Post 302612729 by balajesuri on Monday 26th of March 2012 09:44:13 AM
Old 03-26-2012
If a parameter is given along with a script name while invoking it, then that parameter is considered, otherwise 'today' would contain `date +%y%m%d`.

For e.g., consider this script:
Code:
#! /bin/bash
today=`date '+%y%m%d'`
today=${1:-${today}}
echo $today

While invoking it, if you provide a parameter then variable 'today' would contain that parameter:
Code:
[user@host ~]$ ./test.sh 120331
120331

If you invoked the script without any parameters, then variable 'today' would hold today's date:
Code:
[user@host ~]$ ./test.sh
120326

This User Gave Thanks to balajesuri For This Post:
 
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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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