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Top Forums Shell Programming and Scripting finding the files based on date.. Post 302521781 by apple2685 on Thursday 12th of May 2011 08:43:55 AM
Old 05-12-2011
Bug finding the files based on date..

Hi to every one ,

i had ascenario like this..

i had path like

export/home/pmutv/test/

in this i will recive 43 files daily with each file having that days date
i.e like product.sh.20110512

like this i will 43 files every day

i had to find the files. if files are avaliable i had to send a mail to the user and have to copy the files to another location.

can u people give me the suggestion on this..

thanks in advance..Smilie
regards,
venkat
 

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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)
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