I am new to UNIX and I am trying to write a shell script. I want to be able to list all files that were created with yesterdays dates (APR 29 as an example) that are not 0 file size.Then in those files I want to look for the string 'Process Complete' and list all files that DONT have that string.... (8 Replies)
Hi Guys.
I am very new to UNIX.
I need to get yesterdays and tommorows date given todays date.
Which command and syntax do i use in basic UNIX shell.
Thanks. (2 Replies)
i tried to use "find" to get all of yesterdays files but missed something in the 24 hours logic.
can anybody help me with this one?
i thought that -daystart -atime 1 was enough but i got more files (2 Replies)
Hi Friends,
How to list todays file from a directory listing of files for amny dates.
I tried with the following options but not working :
find . -name "esi01v*" -mtime 1 -ls
find . -name "esi01v*" -ctime 1 -ls
find . -name "esi01v*" -mtime 1
Please advise (19 Replies)
I will be very grateful if someone can help me with bash shell script that does the following:
I have a list of filenames:
A01_155716
A05_155780
A07_155812
A09_155844
A11_155876
that are kept in different sub directories within my current directory. I want to find these files and copy... (3 Replies)
Hi All,
I am mediator Shell programmer, Just have an hands on experice :-), i am writing a shell scirpt to list logs of todays date from /var/log/messages.
I need to ur kind help where if i run this script from cron. the script should filter todays logs only from /var/log/messages.
Below... (4 Replies)
Hi all,
i have a folder, with tons of files containing as following,
on /my/folder/jobs/
some_name_2016-01-17-22-38-58_some name_0_0.zip.done
some_name_2016-01-17-22-40-30_some name_0_0.zip.done
some_name_2016-01-17-22-48-50_some name_0_0.zip.done
and these can be lots of similar files,... (6 Replies)
Hello,
I have a syslog server at home and am currently experiencing an issue where my logs will rotate and compress however it will rotate and compress yesterdays file and the newly created log file for the current day. When it does this however it will also create another new file for today... (9 Replies)
Discussion started by: MyUserName7000
9 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)