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Top Forums Shell Programming and Scripting Help File Copy Historical Files Post 302532898 by srattani on Wednesday 22nd of June 2011 10:24:07 AM
Old 06-22-2011
Hello,

Thanks, but I do not follow. Date calc script? First I guess some how I need to get the date out from the file name isnt it? I am not quite sure how and then some how compare that date with todays date to see if days in between is less than or equal to 300 then copy over the file.

I did not quite understand the oracle part, I would like to accomplish it all in KSH.

Would appreciate any help
 

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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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