I need to extract the date part from the file name (20080221 in this ex) and compare it with the current date and delete it, if it is a past date.
$file = exp_ABCD4_T-2584780_upto_20080221.dmp.Z
really appreciate any help.
thanks
mkneni (4 Replies)
Hey guy,
how to make bash script to create foo.txt file and add current date into file content and that file always append.
example: today the script run and add today date into content foo.txt
and tomorrow the script will run and add tomorrow date in content foo.txt without remove today... (3 Replies)
Hi Gurus,
I'm using HP-UX B.11.23 operating system.
I've been trying to extract this log info based on the current date and month, but was having some issues as the date column which on the 4th column has a comma and the 5th column has a dot tied to it.
Here is the output from my shut... (5 Replies)
Hello gurus,
I am hoping someone can help me with the required code/script to make this work. I have the following file with records starting at line 4:
NETW~US60~000000000013220694~002~~IT~USD~2.24~20110201~99991231~01~01~20101104~... (4 Replies)
Hi,
I am trying to display future date from the current date but unable to do so in UNIX (not in PERL). For eg: if today is March 5 then I want a variable wherein I can store Mar 7 date, but unable to get the future date from the current date.
I have tried many possible ways as mentioned below... (11 Replies)
Hi all,
Following is my small script:-
#!/bin/ksh
for i in `cat /users/jack/mainfile-dr.txt`
do
sudo cp -r $i /users/jack/DR01/.
done
cd /users/jack/DR01/
sudo tar cvf system1-DR.tar *
scp system1-DR.tar backupserver:/DRFiles/system1
sudo rm -rf system1-DR.tar
In this script I... (10 Replies)
i have file 1.txt
asdas|csada|13|03|10|04|23|A1|canberra
sdasd|sfdsf|13|04|26|23|28|A1|sydney
i want to add today's date and time in the end of each row
expected output
asdas|csada|13|03|10|04|23|A1|canberra|130430|1358
sdasd|sfdsf|13|04|26|23|28|A1|sydney|130430|1358
todays date... (10 Replies)
SunOS -s 5.10 Generic_147440-04 sun4u sparc SUNW,SPARC-Enterprise
Hi,
In a folder, there are files. I have a script which reads the current date and subtract the modification date of each file.
How do I achieve this?
Regards,
Joe (2 Replies)
We want to call a parameter file (.txt) where my application read dynamic values when the job is triggered, one of such values are below:
abc.txt
------------------
Code:
line1
line2
line3
$$EDWS_DATE_INSERT=08-27-2019
line4
$$EDWS_PREV_DATE_INSERT=08-26-2019
I am trying to write a... (3 Replies)
Discussion started by: pradeepp
3 Replies
LEARN ABOUT PHP
datetime.add
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)