Hi
I am having date as a string in DDMMYYYY format(07082008) in a variable say cdate. I want to Convert it into DD Month YYYY format(7 August 2008). Could someone help. Thanks in Advance. (2 Replies)
I have a file which has 100k+ records like this
abc,05-JUN-1974,def,lkj,aaa
def,11-SEP-1975,ghj,dis,dea
I want to convert ex 05-JUN-1974 to 06/05/1974
Please help me with awk script to convert the whole file into MM-DD-YYYY
Thank you! (2 Replies)
In my shell script i have a variable which stores date in the format of YYYYMMDD. Is there any way to format this value to MM/DD/YYYY.
Thanks. (8 Replies)
Hi all
I have some pipe-separated data in the form:
5/12/2008 00:00:00|31/1/2009 00:00:00|SOMESTUFF|OTHERSTUFF
12/31/2008 00:00:00|15/1/2009 00:00:00|MORESTUFF|REMAININGSTUFF
1/1/1023 00:00:00|16/5/2047 00:00:00|THEREST|YETMORE
I need to zero-pad the single-digit days and months, using... (3 Replies)
My csv has data like this
x,x,3452,2/18/1986,abc
x,g,19711,1/24/1986,abc
i want to replace date in the following format YYYY-mm-dd
how do i convert using awk script ? (8 Replies)
How can I convert any user inputted date into yyyy/mm/dd ?
For example user can input date one of the following 20120121 , 2012-01-21 ,01/21/2012,01/21/2012 etc
But I need to convert any of the date entered by user into yyyy/mm/dd (2012/01/2012). Any suggestion. Thanks in advance
this is... (1 Reply)
Hi All,
I have file like
“April 10, 2013”,”raj”
“April 29, 2013”,”raj1”
Output :
“2013/04/10”,”raj”
“2013/04/29”,”raj1”
Please help me how to do... (9 Replies)
Hello All,
I have a date in DD/MM/YYYY format. I am trying to convert this into unix timestamp. I have tried following:
date -d $mydate +%s
where mydate = 23/12/2016 00:00:00
I am getting following error:
date: extra operand `+%s'
Try `date --help' for more information.
... (1 Reply)
How to convert mmm-yy to mm/dd/yyyy format in unix ?
example:
Jan-99 to 01/01/1999
Jan-00 to 01/01/2000
Jan-25 to 01/01/2025
Dec-99 to 01/12/1999
Dec-00 to 01/12/2000
Dec-25 to 01/12/2025
YY anything between 00-50 should be 2000-2050
YY anything between 51-99 should be 1951-1999
... (2 Replies)
Discussion started by: gksenthilkumar
2 Replies
LEARN ABOUT LINUX
bashbug
BASHBUG(1) bashbug BASHBUG(1)NAME
bashbug - report a bug in bash
SYNOPSIS
bashbug [--help] [--version] [bug-report-email-addresses]
DESCRIPTION
bashbug is a utility for reporting bugs in Bash to the maintainers.
bashbug will start up your preferred editor with a preformatted bug report template for you to fill in. Save the file and quit the editor
once you have completed the missing fields. bashbug will notify you of any problems with the report and ask for confirmation before
sending it. By default the bug report is mailed to both the GNU developers and the Debian Bash maintainers. The recipients can be changed
by giving a comma separated list of bug-report-email-addresses.
If you invoke bashbug by accident, just quit your editor. You will always be asked for confirmation before a bug report is sent.
OPTIONS --help
Show a brief usage message and exit.
--version
Show the version of bashbug and exit.
bug-report-email-addresses
Comma separated list of recipients' email addresses. By default the report is mailed to both the GNU developers and the Debian Bash
maintainers.
ENVIRONMENT
DEFEDITOR
Editor to use for editing the bug report.
EDITOR
Editor to use for editing the bug report (overridden by DEFEDITOR).
SEE ALSO bash(1), reportbug(1), update-alternatives(8) for preferred editor.
AUTHOR
This manual page was written by Christer Andersson <klamm@comhem.se> for the Debian project (but may be used by others).
GNU Bash 3.1 11 December 2007 BASHBUG(1)