I have a shell script which gets passed a parameter which is a combination of Year and Julian Date <YYYYj>. So April 11th, julian date is 101. So if I wanted April 11th for 2003 I would get the following value 2003101. How would I convert that in unix to be 20030411? I am using the korn shell. (3 Replies)
I need Unix, ksh scripts that will convert dates - Gregorian to Julian and Julian to Gregorian.
Input for converting Gregorian to Julian would be in the form of CCYYMMDD with the output being CCYYDDD.
Input for converting Julian to Gregorian would be CCYYDDD with the output being CCYYMMDD.
... (4 Replies)
Hi,
is there any possibility to find julian date for given corresping date.
I will be gladfull if i get it.
Requirement :
Input : 10 09 2006
output: julian date: 283
thanks
srikanth (2 Replies)
Hi,
I have script in unix which creates a julian date like 126 or 127
I want convert this julian date into calender date
ex : input 127
output 07/may/2007 or 07/05/2007 or 07/05/07
rgds
srikanth (6 Replies)
Hi,
im new for UNIX. i have a problem in date function. please help me to find a solution.
batchdate="29/10/2010"
nextdate="01/11/2010"
i want compare this two date. if my batch date greater than nextdate should prompt error message. how can i do that? as i know its better and safer if i... (2 Replies)
Use and complete the template provided. The entire template must be completed. If you don't, your post may be deleted!
1. The problem statement, all variables and given/known data:
This function is given the day, month and year and returns the Julian date. The Julian date is the... (1 Reply)
All, I am facing an issue with julian date conversion..
current command:
echo `date +%Y%j` `cat -n /home/user/FILENAME.dat |awk '{printf "%08s", $2}'`
The above command is working good. But in the above bolded part, it is converting system date to julian date. However I want to... (8 Replies)
Need assistance . Below code gets me julian date . I wanted to add hour/24 to julian date and output it. Is there a way to do the calculation?
use Time::Local;
use POSIX qw(strftime);
my $time=timelocal(1,2,3,9,11,2013);
printf strftime "%j", localtime($time);
343 (3 Replies)
How to get Julian date (Three digit) of a given date (Not current date)? I do not have root privilege - so can not use date -d. Assume that we have three variables year, month and date.
Thx (5 Replies)
Discussion started by: Soham
5 Replies
LEARN ABOUT PHP
datetime.add
DATETIME.ADD(3) 1 DATETIME.ADD(3)DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::add (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_add (DateTime $object, DateInterval $interval)
Adds the specified DateInterval object to the specified DateTime object.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $interval
- A DateInterval object
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
EXAMPLES
Example #1
DateTime.add(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-11
Example #2
Further DateTime.add(3) examples
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-01 10:00:30
2007-06-05 04:03:02
Example #3
Beware when adding months
<?php
$date = new DateTime('2000-12-31');
$interval = new DateInterval('P1M');
$date->add($interval);
echo $date->format('Y-m-d') . "
";
$date->add($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.sub(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.ADD(3)