:) I want to rename a file by giving the actual date as a file name.
Is there a command for this?
For example file is this: a.txt
I want to copy it with a command like this: a20050510.txt (3 Replies)
I am a newbie in shell scripting, hope you can advice how can I compare the date/time of file extracted from 'll' and current system date/time.
I have done the following:
ll -rt > $FILE_AGE_LOG
FILE_DATETIME=`more $FILE_AGE_LOG | head -02 | cut -c 45-57`
It returns 'May 4 19:11'.
If I... (4 Replies)
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.
$file = exp_ABCD4_T-2584780_upto_20080221.dmp.Z
really appreciate any help.
thanks
mkneni (4 Replies)
Hi,
I'm using the ksh shell and I'd like to set my PS1 prompt on an AIX system to include, amongst ther things, the current time.
This was my best effort: export PS1=$(date -u +%R)'${ME}:${PWD}# '
but this only sets the time to the value when PS1 is defined and the time value doesn't... (4 Replies)
I am trying to compile a file called PPFormatageMUT.c in which I have included header file which are at some other location but the point is that while compiling the file, it is throwing error saying that
error : no such file or directory
source code location:... (1 Reply)
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 want to put script. The script is to show file that larger than 100MB include current date (eg: today date).
After find the date, it will compress list file and split the tar.gz file. Any idea how to do that?
This bash script will run auto everyday. It's will transfer will to other... (12 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)
I Have text like
XXX_20190908.csv.gz need to replace Only date in this format with current date every day
Thanks! (1 Reply)
Discussion started by: yamasani1991
1 Replies
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)