02-28-2002
jobs
I am running this script below as a cron job for user root. In fact I set this job early last week and expected it to delete archive logs older than 7 days, but doesn't work - just displays the echo messages and no error. Can somebody advise me the probable causes ??
DAY_AFTER=7; export DAY_AFTER
DELETE_FROM_DIR=/u05/oracle/oradata/TOLL2/archlog; export DELETE_FROM_DIR
date
echo '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'
echo 'About to delete archive logs...'
echo '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'
find $DELETE_FROM_DIR -name "*.*" -mtime +$DAY_AFTER -exec rm {} \;
echo '~~~~~~~~~~~~~~~~~'
echo 'Process Complete.'
echo '~~~~~~~~~~~~~~~~~'
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)