I am writing a script that parses different logs and produces one. In the source files, the date is in DD MM YYYY HH24:MI:SS format. In the output, it should be in DD MON YYY HH24:MI:SS (ie 25 Jan 2010 16:10:10)
To extract the dates, I am using shell substrings, i.e.:
Code:
read line
year=${line:8:4}
month=${line:5:2}
day=${line:2:2}
hour=${line:13:2}
min=${line:16:2}
sec=${line:19:2}
echo "$year$month$day$hour$min$sec" # prints something like 20110325161610 (I use this numeric value to compare it agains reference date, i.e. date when this script ran last time)
if [ "$year$month$day$hour$min$sec" -gt "$lastlog" ]
then
# output date in DD MON FORMAT... how?
fi
Of course I can do a routine like: if mon = 01 then mon = 'JAN', if mon = 02 then mon = 'FEB'... But I guess there must be a nicer solution to this. Maybe solution that will take into account also LANG setting (not a requirement, but would like to have it).
Maybe date function can be used for that, but I did not figure out how to pass the -d parameter (what format should it be in?). Did not find much in FAQ: Date Arithmetics either
I want to change a date from format dd-mmm-yyyy to mm/dd/yyyy. Is there a way to do this with sed or do you have to write a case statement to convert JAN to 01? Thanks (9 Replies)
How to convert the date field from dd/mm/yyyy to yyyy/mm/dd in unix
my script will generate text file which have two fields
one is date and another is name of the server for example this is sample date which I have to sort based on older to newer date the problem is when I found out sort will... (4 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)
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)
Hi All,
I am learning PERL for one of the projects, and in one of these scripts, I read a flat text file and print in the terminal.
The problem is, the text file has a date field. The format is yyyymmdd. I need to display this as dd-mon-yyyy.
Any ideas to do this? Thanks a lot for the... (9 Replies)
(Attention: Green PHP newbie !)
I have an online inquiry form, delivering a date in the form yyyy/mm/dd to my feedback form. If the content passes several checks, the form sends an e-mail to me. All works fine. I just would like to receive the date in the form dd/mm/yyyy. I tried with some code,... (6 Replies)
Hi I have a problem with Date format in my code.
1st I am trying to convert today's date to yesterday's using
YESTERDAY3=`perl -e '@y=localtime(time()-86400); printf "%04d/%02d/%02d",$y+1900,$y+1,$y;$y;'`
And once it is done I am trying to using the yesterday date in a grep command to... (3 Replies)
I've seen a lot of posts on this and have tried the following:
echo 1257000000| perl -e '($d,$m,$y)=(localtime(time-86400));$m+=1;$y+=1900;printf "$y/$m/$d\n";'
But I am unable to convert a past Epoch date into a format such as YYYY/MM/DD or MM/DD/YYYY.
I am using bash and don't know... (4 Replies)
I am getting output of YYYY-MM-DD and want to change this to DD/MM/YYYY.
When am running the query in 'Todd' to_date(column_name,'DD/MM/YYYY') am getting the required o/p of DD/MM/YYYY, But when am executing the same query(Netezza) in linux server(bash) am getting the output of YYYY-MM-DD
file... (3 Replies)
Discussion started by: Roozo
3 Replies
LEARN ABOUT SUSE
grep-changelog
grep-changelog(1) General Commands Manual grep-changelog(1)NAME
grep-changelog - print ChangeLog entries matching criteria
SYNOPSIS
grep-changelog [options] [CHANGELOG...]
DESCRIPTION
grep-changelog searches the named CHANGELOGs (by default files matching the regular expressions ChangeLog and ChangeLog.[0-9]+) for
entries matching the specified criteria. At least one option or file must be specified. This program is distributed with GNU Emacs.
OPTIONS
The program accepts unambiguous abbreviations for option names.
--author=AUTHOR
Print entries whose author matches regular expression AUTHOR.
--text=TEXT
Print entries whose text matches regular expression TEXT.
--exclude=TEXT
Exclude entries matching regular expression TEXT.
--from-date=YYYY-MM-DD
Only consider entries made on or after the given date. ChangeLog date entries not in the "YYYY-MM-DD" format are never matched.
--to-date=YYYY-MM-DD
Only consider entries made on or before the given date.
--rcs-log
Print output in a format suitable for RCS log entries. This format removes author lines, leading spaces, and file names.
--with-date
In RCS log format, print short dates.
--reverse
Show matches in reverse order.
--version
Display version information.
--help Display basic usage information.
COPYING
Copyright (C) 2008, 2009 Free Software Foundation, Inc.
Permission is granted to make and distribute verbatim copies of this document provided the copyright notice and this permission notice are
preserved on all copies.
Permission is granted to copy and distribute modified versions of this document under the conditions for verbatim copying, provided that
the entire resulting derived work is distributed under the terms of a permission notice identical to this one.
Permission is granted to copy and distribute translations of this document into another language, under the above conditions for modified
versions, except that this permission notice may be stated in a translation approved by the Free Software Foundation.
grep-changelog(1)
04-05-2011
