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Top Forums Shell Programming and Scripting reading a file extracting information writing to a file Post 302563895 by vbe on Wednesday 12th of October 2011 11:08:38 AM
Old 10-12-2011
At the first line:
Code:
 today=`date +%c%m%d | cut -c24-31`
 echo ${today}

what should it display?
When I try (OK AIX 4.2.1...) I get:
Code:
page:/etc $  today=`date +%c%m%d | cut -c24-31`
page:/etc $  echo ${today}
11012

Is that what you want? (I would have expected 111012...)
If its 111012 then why not simply:
Code:
page:/etc $ TODAY=$(date "+%y%m%d")
page:/etc $ echo $TODAY
111012

Ah Im using ksh...
 

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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)
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