I need to determine if any three files have the same file size in a specified directly?
I have got as far as listing the file sizes but where to go from here?
ls -al |sort -n -r +4 | awq '{print $5}'
Thanks in anticipation (5 Replies)
I need to zip all the files within a directory on a daily basis and store them as a zip file within that directory with the date in its naming convention
I have file extentions .log .lst and .out's within that directory
How do i do this task
zip -mqj mydir/allzip$YYMMDDDD.zip mydir/*.*
... (5 Replies)
We recieve some logs on our windows box via FTP on a daily basis, in the same directory. I would like to check for missing logs files and add their name to a text file.
Something like...
Check if C:\logs\file1_currentdate exists (if not, add file1_currentdate to... (1 Reply)
Need help on this script.
the month is not changing to February...
#!/bin/bash
for X in `seq 1 100`
do
DATE=`date +%Y-%m-%d "--date=${X} day ago"`
Y=`date +%Y`
M=`date +%m`
D=`date +%d "--date=${X} day ago"`
DIR=/home/LogBackup
for i in `seq 1 `
do
if ;then
# ... (1 Reply)
Hi
I am still learning how to write shell scripts, so I started to write a script like this:
#!/bin/sh
date
echo
outputOK () {
echo $1 ""
}
outputOK () {
echo $1 ""
}
for vol in `/usr/bin/grep -E 'hfs|vxfs|nfs|cifs' /etc/fstab | egrep -v '^#' | awk '{ print $3 }'`
do
if... (7 Replies)
hi,
I am having script in which i want to check if directory has any file in it or not. If directory contains a single or more files then and only then it should proceed to further operations...
P.S.: Directory might have thousand number of files. so there might be chance of getting error... (4 Replies)
I'm working on a bash script to move files from one location, to two. The first part of my challenge is intended to check a particular directory for contents (e.g. files or other items in it), if files exists, then send the list of names to a txt file and email me the text file. If files do not... (4 Replies)
Is there a way to calculate the total file size of HDFS file directory in GB or MB? I dont want to use du/df command. Without that is there a way
HDFS
Directory - /test/my_dir (1 Reply)
I'm trying to wirte ksh script for given requirement, but i unable to achive it.
In dir1 directory I need to check for the files which suffixed with .csv or .txt, If there is no files, then i need to exit. If any files found I need to move the each file found to dir2 directory. I have to repeat... (4 Replies)
Discussion started by: Kayal
4 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)