Sponsored Content
Top Forums Shell Programming and Scripting How to change date format in 'last' command? Post 302519125 by thearpit on Tuesday 3rd of May 2011 05:22:07 AM
Old 05-03-2011
hi
thanks or the reply..but my question is- is there any switch avaliable for the last command to change the date format.

actually the real problem is --

file 'temp' contains the data of last command.
now i am reading the file through while loop. and want to display only those lines which have date >= todays' date.
i want to do this by using awk. plese tell if you have any idea on this.

thanks
 

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

How to change it to the date format

Hi, I want to know how to change this string to date format 20061102122042 to 02-11-2006 12:20:42 or 02-Nov-2006 12:20:42 Please let me know at the earliest.Thanks in advance. Regards, Preetham R. (3 Replies)
Discussion started by: preethgideon
3 Replies

2. Shell Programming and Scripting

Change of date format

I want to chnage the date format from the file format like below to WT;T15D;0000007208;;20080401;3;0;0;3;;B;ZZZZZZ; WT;T25D;0000007208;;20080401;6;0;0;6;;B;ZZZZZZ; WT;T5D;0000007208;;20080401;123;0;0;123;;B;ZZZZZZ; to WT;T15D;0000007208;;04/01/200804;3;0;0;3;;B;ZZZZZZ;... (2 Replies)
Discussion started by: svenkatareddy
2 Replies

3. Solaris

change date format

dear members, ls -l drwxr-xr-x 40 root sys 1024 Jul 11 22:19 usr drwxr-xr-x 43 root sys 1024 Feb 1 2009 var i am using solaris 10 is that possibe to do drwxr-xr-x 40 root sys 1024 25-08-2009 22:19 usr drwxr-xr-x 43 root sys ... (4 Replies)
Discussion started by: hosney00ux
4 Replies

4. Shell Programming and Scripting

Date Format Change

Hi I have a date format in a variable as Mon Jan 11 03:35:59 2010. how do i change it to YYYYMMDD format (3 Replies)
Discussion started by: keerthan
3 Replies

5. Shell Programming and Scripting

Change date format

Hi guys, I have a text file with lots of lines like this: MCOGT23R27815 27/07/07 27/05/09 SO733AM0235 30/11/07 30/11/10 NL123403N 04/03/08 04/03/11 0747AM7474 04/04/08 04/04/11 I want to change each line so the date format looks like this: MCOGT23R27815 07/07/27 09/05/27 ... (7 Replies)
Discussion started by: Tornado
7 Replies

6. Shell Programming and Scripting

Format of SED command to change a date

I have a website. I have a directory within it with over a hundred .html files. I need to change a date within every file. I don't have an easy way to find/replace. I need to change 10/31 to 11/30 on every single page at once. I tried the command below but it didn't work. Obviously I don't know... (3 Replies)
Discussion started by: ijustsawmars
3 Replies

7. Shell Programming and Scripting

Change Date Format

Hi Guys, I had a scenario like this.. It seems very silly...dont think it as a home work question.....:) i tried it many ways but i didn't achieve this... start_date=May122011 here i want to change the start_date in to 20110512 start_date=20110512 tell me how can we achive... (5 Replies)
Discussion started by: apple2685
5 Replies

8. Shell Programming and Scripting

How to change the date format

Hi Guys, Can someone help me on how to change the date format using sed or any unix command to give my desired output as shown below. INPUT FILE: 69372,200,20100122T17:56:02,2 53329,500,20100121T11:50:07,2 48865,100,20100114T16:08:16,2 11719,200,20100108T13:32:20,2 DESIRED... (2 Replies)
Discussion started by: pinpe
2 Replies

9. UNIX for Dummies Questions & Answers

Change Date from one format to other

Hi I wish to change date from one format to another in unix. eg: INPUT DATE: 2013159 (YEAR & NUMBER OF DAY) OUTPUT DATE required: 20130608 (YYYYMMDD) how to do it ? Thanks in advance. (6 Replies)
Discussion started by: dashing201
6 Replies

10. Shell Programming and Scripting

Change just the date format

need some more help please i have a large file and i want to change just the date format with awk or sed from this yyyy-mm-dd 2016-04-15 8.30 2016-04-15 7.30 2016-04-13 7.30 2016-04-11 8.30 2016-04-11 7.30 2016-04-08 8.30 2016-04-08 7.30 2016-04-06... (5 Replies)
Discussion started by: bob123
5 Replies
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)
All times are GMT -4. The time now is 07:10 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy