hi all,
i had a requirement in my script iam passing the parameter like (its not the system date iam passing as argument like it can be any date)
but in my output file it should be
please guide me
I have a comma delimited log file which has the date as MM/DD/YY in the 2nd column, and HH:MM:SS in the 3rd column.
I need to change the date format to YYYY-MM-DD and merge it with the the time HH:MM:SS. How will I got about this?
Sample input
02/27/09,23:52:31
02/27/09,23:52:52... (3 Replies)
Hi
According with the 'date' command documentation
My question is, which are the possible values for STRING? I have been looking in the man pages but I didn't find anything?
After some googling I have found it can have special values as 'now', 'yesterday' or 'tomorrow'. If it represents... (1 Reply)
Hi,
Unix: HP-UNIX.
I have a requirement(part of my requirement) to get latest file modification date of all files(In a particular directory).
Using ll comand i m able to find out the modification date but not in the required format
My command and out put:
echo trp_pb.sql `ll... (2 Replies)
hi..
i am new to here. can anybody tell me how can we change the date format in the 'last' command.
EX- on running last command i am getting --
rruat pts/12 172.18.40.101 Tue May 3 12:59 still logged in
rruat pts/10 blr2-3f-239.asco Tue May 3 12:59 - 13:09 ... (2 Replies)
Hi...
I have an inputfile name as :- abc_test_20120213.dat (date in yyyymmdd format)
I need the output file name as abc_test_13022012.dat (date in ddmmyyyy format)
Please help me on this...
Thanks in advance. (5 Replies)
Hi All,
Following is my issue.
$MAIL_DOC = test.txt
test.txt contains the following text .
This process was executed in the %INSTANCE% instance on %RUNDATE%.
I am trying to execute the following script
var=`echo $ORACLE_SID | tr `
NOW=$(date +"%D")
sed -e... (3 Replies)
Hi Unix Gurus,
I would like to rename several files in a Unix Directory . The filenames can have more than 1 underscore ( _ ) and the last underscore is always followed by a date in the format mmddyyyy. The Extension of the files can be .txt or .pdf or .xls etc and is case insensitive ie... (1 Reply)
Hi,
I need to convert a date format "August 12, 2013 9:40:00 PM CDT" in to DD/MM formant
For example ..I am using ...
date -d "January10, 2013 04:05:00 AM CST" +%m/%d
which gives me...
01/10
However, when i'am using it against every month it is throwing errors on some months as... (3 Replies)
I have a script called " passwd_status " which gives the passwd-s status of my servers.
the script output is like below
password status for A:
abc ks 10/05/115 1 30 ps
password status for B:
abc ks 09/25/115 1 30 ps
password status for C:
abc ks 10/10/115 1 30 ps
Now , i want to... (1 Reply)
Record:
Record1|Record2|Record3|Record4|Record5|DATE1|DATE2
Need to Check DATE1 & DATE2 is in DDMMYYYY format in a file.
records which not meet the date format DDMMYYYY extract to other file. (1 Reply)
Discussion started by: vivekn
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)