08-16-2001
Date Handling
In the shell which have the command prompt '$' ( Sorry i dunno wat shell is this), is there anyway to handle date input?
Coz i need to accept a date from the user and wat i use is:
read day;read month; read year;
In this case user need to enter 3 time, is it any better way?
Summore, is there any problem with this command??
checkCategory() {
case $category in
"Soft"|"Hard") return 1 ;;
*) echo "error!"
return 0 ;;
esac
}
while [ ' checkCategory $category ' ]
do
read category
done
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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)