10-10-2006
Hi srikanthus2002, have you searched the forums for this? Try searching for 'date calculation' or 'date arithmetic'. You will find stuff that will help you get the answer.
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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)
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Hi,
I have script in unix which creates a julian date like 126 or 127
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output 07/may/2007 or 07/05/2007 or 07/05/07
rgds
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Hi,
Was using date +%Y%j to get current julian date. Can anyone let me know how can I get y'day's julin date. Thx
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Hi Gurus,
Need help in Conversion of date(2007-11-30) to Julian date(YYDDD)...
'+%J'
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Hi,
im new for UNIX. i have a problem in date function. please help me to find a solution.
batchdate="29/10/2010"
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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)
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Hi all,
I require to convert julian date to normal calander date in unix
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----------------------------------------
gr8 if give very small command/script
and please explain the steps as well(imp)
Thanks
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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);
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printf strftime "%j", localtime($time);
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10. Shell Programming and Scripting
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.
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LEARN ABOUT PHP
datetime.sub
DATETIME.SUB(3) 1 DATETIME.SUB(3)
DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_sub (DateTime $object, DateInterval $interval)
Subtracts the specified DateInterval object from 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.sub(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-10
Example #2
Further DateTime.sub(3) examples
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-19 13:59:30
1992-08-15 19:56:58
Example #3
Beware when subtracting months
<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-03-30
2001-03-02
NOTES
DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO
DateTime.add(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.SUB(3)