I'm trying to rename the last generated file in a given directory using a makefile.
If I type 'make build' the 'build' target produces a file (ie: text_001.txt). It automatically generates them sequentially, so I don't have control over the name. I can't say that it will always be text_001.txt.... (2 Replies)
Hi,
Can anyone plz share their experience with -
Building shell script to append the file with date in following format-
Filename_MMDDYYYY.txt
Thanks in advance (2 Replies)
Hi I am very new to scripting,
Can someone show me how to (in unix shell script) compare the system's date with a date in a file. The requirement is to somehow open this file (which will only have a date in it) and compare it with today's date. If they are equal execute a procedure below but if... (4 Replies)
I am a newbie to scripting.
I need a korn shell script to copy log files of current day to archive folder and rename with current days date stamp.
I would really appreciate your help.
File structure is as follows. Everyday files get overwritten, so I need copy to a archive directory and... (3 Replies)
Hi Guys,
I was working some time ago n was in need to calculate date 30/31 days from today including Feb (Leap yr stuff). Today date is variable depending on day of execution of script. I tried searching but was not able to get exactly what I needed....So at that I time I implemented by my own... (3 Replies)
I would like to find the Files which are generated today in the current directory:
I use the commad ls -lrt * | egrep " `date "+%b"` * `date "+%d"`
to acheive this. Is there any better way to acquire the same.
Multiple answers will be great. Thanks (3 Replies)
Hi
I have file with number status and date1 and date1 field,
want add a column today between column date1 and date2.
file1.txt
number status date1 date2
===== ==== === =====
34567 open 27/06/13 28/06/13
45678 open 27/06/13 28/06/13
43567 open 27/06/13 28/06/13 ... (1 Reply)
This is on a CentOS box, I have two scripts that need to run in order.
I want to write a shell script that calls the first script, lets it run and then terminates it after a certain number of hours (that I specify of course), and then calls the second script (they can't run simultaneously) which... (3 Replies)
Hi Community!
Following on from this code in another thread:
#!/bin/bash
file_string=`/bin/cat date.txt | /usr/bin/awk '{print $5,$4,$7,$6,$8}'`
file_date=`/bin/date -d "$file_string"`
file_epoch=`/bin/date -d "$file_string" +%s`
now_epoch=`/bin/date +%s`
if
then
#let... (2 Replies)
Discussion started by: Greenage
2 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)