Date: Thu, 13 Jan 2011 06:28:19 +0100
From: Polytropon free...@edvax.de
Subject: awk question: replacing %d%s by %d %s
I'm aware that this is not an awk question list, but I'm confident there
are many awk gurus here who can surely help me with such a stupid
problem. I also know that I
On Thu, 13 Jan 2011 18:22:18 -0600 (CST), Robert Bonomi
bon...@mail.r-bonomi.com wrote:
True. But
sub(nr,[a-z], );
does the trick. (tested on Freebsd 7.2)
Explamation: is a 'replacement side' magic incantation to the regex
library that means 'that which was matched by the
On Fri, 2011-01-14 at 07:17 +0100, Polytropon wrote:
On Thu, 13 Jan 2011 18:22:18 -0600 (CST), Robert Bonomi
bon...@mail.r-bonomi.com wrote:
True. But
sub(nr,[a-z], );
does the trick. (tested on Freebsd 7.2)
Explamation: is a 'replacement side' magic incantation to the
On Fri, 14 Jan 2011 17:53:04 +1030, Wayne Sierke w...@au.dyndns.ws wrote:
I suspect it is a transcription error by Robert in his email.
From man awk:
sub(r, t, s)
substitutes t for the first occurrence of the regular
expression
r in the string s. If s
On Thu, 13 Jan 2011 01:00:17 -0500, Tom Limoncelli t...@whatexit.org wrote:
$ awk data.txt experiment.txt '{ num = $1 ; sub(/[^0-9]+$/, ,
num) ; lets = $1 ; sub(/^[0-9]+/, , lets); print num lets }' ;
diff -cw control.txt experiment.txt
$ # The above puts a space at the end of the first 3
On Thu, Jan 13, 2011 at 12:28 AM, Polytropon free...@edvax.de wrote:
I have strings of the form either number(s) or
number(s)letter. I catch them with
...
where nr is the name of the string. What I need
is a simple space between number(s) and letter,
so for example 12a would get 12 a, 6d
On Wed, 22 Apr 2009 12:38:47 -0700, Evuraan::ഏവൂരാന് evur...@gmail.com
wrote:
but this below, does not work
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { print
$0 | mail m...@email.address }'
I would suggest to keep the system() approach:
tail -f
You might want to look at ``swatch'' which is designed to do
this, and monitors multiple log files simultaneously.
On Wed, Apr 22, 2009, Evuraan:: wrote:
Greetings..!
this works,
tail -f /var/log/apache2/access.log | nawk '/192.168.1.100/{ print $0 }'
and this too:
nevermind, i got it to work, with a little help from
http://student.northpark.edu/pemente/awk/awk_sys.txt,
tail -f /var/log/apache2/access.log | awk '/192.168.1.100/
{system(echo $0 | mailx -s test_email m...@email.com ) }'
thx..!
2009/4/22 Bill Campbell free...@celestial.com:
You might want
On Thu, 9 Apr 2009 15:32:51 +0200 (CEST), Oliver Fromme
o...@lurza.secnetix.de wrote:
If ; is the delimiter character, you need to tell awk
about it (i.e. use the -F option). This one should work:
awk -F';' '$3 ~ /^[a-z]{5}$/ {print}' file
You can even omit {print} because it's the
Just add a filter
NF 2
to the script. You can even take care of 1 token lines and empty lines
in whatever way you wish with other filters.
___
freebsd-questions@freebsd.org mailing list
awk '{print $(NF-1)}' user.csv
Yup, those blank lines will kill it for sure. A sed filter to
remove blank lines ahead of the awk statement should allow it to
work properly.
Or awk only i.e. no sed:
awk '!(/^$/) { print $(NF-1) }' user.csv
--
Nino
n j writes:
Or awk only i.e. no sed:
awk '!(/^$/) { print $(NF-1) }' user.csv
That's right. I originally suggested the sed and then was
thinking about it as I walked home yesterday and knew that awk
could test for the blank line condition before committing
suicide.:-)
Martin
P.U.Kruppa wrote:
Hi (and sorry for this slightly OT question),
I would like to extract the second last field of each line of a file
called user.csv .
So I try
awk '{print $(NF-1)}' user.csv
awk: trying to access out of range field -1
input record number 1, file user.csv
On Thursday 26 July 2007 15:26:02 Peter Boosten wrote:
P.U.Kruppa wrote:
Hi (and sorry for this slightly OT question),
I would like to extract the second last field of each line of a file
called user.csv .
So I try
awk '{print $(NF-1)}' user.csv
awk: trying to access
On Thu, 26 Jul 2007, Peter Boosten wrote:
P.U.Kruppa wrote:
Hi (and sorry for this slightly OT question),
I would like to extract the second last field of each line of a file
called user.csv .
So I try
awk '{print $(NF-1)}' user.csv
awk: trying to access out of range field -1
Don Hinton writes:
On Thursday 26 July 2007 15:26:02 Peter Boosten wrote:
P.U.Kruppa wrote:
awk '{print $(NF-1)}' user.csv
Yup, those blank lines will kill it for sure. A sed filter to
remove blank lines ahead of the awk statement should allow it to
work properly.
Martin McCormick
At 07:43 PM 4/10/2007, Gary Kline wrote:
On Tue, Apr 10, 2007 at 06:35:33PM -0500, Derek Ragona wrote:
At 06:17 PM 4/10/2007, Gary Kline wrote:
On Mon, Apr 09, 2007 at 06:54:07PM -0700, Rick Olson wrote:
I'm assuming you've already taken care of this, but to answer your
original question
On Mon, Apr 09, 2007 at 06:54:07PM -0700, Rick Olson wrote:
I'm assuming you've already taken care of this, but to answer your
original question in AWK form, you could have done the following:
ls -l | awk '$8 == 2006 {system(rm $9)}'
i'Ll save your snippet to my growing %%% awk
At 06:17 PM 4/10/2007, Gary Kline wrote:
On Mon, Apr 09, 2007 at 06:54:07PM -0700, Rick Olson wrote:
I'm assuming you've already taken care of this, but to answer your
original question in AWK form, you could have done the following:
ls -l | awk '$8 == 2006 {system(rm $9)}'
i'Ll
On Tue, Apr 10, 2007 at 06:35:33PM -0500, Derek Ragona wrote:
At 06:17 PM 4/10/2007, Gary Kline wrote:
On Mon, Apr 09, 2007 at 06:54:07PM -0700, Rick Olson wrote:
I'm assuming you've already taken care of this, but to answer your
original question in AWK form, you could have done the
You are trying to remove the files whose names are given by
ls -lt | awk '{if ($8 == 2006) print $9}';
If you are in the same directory, or you have full pathnames, you can
do just (and avoid the 'for do done' loop)
rm $( ls -lt | awk '{if ($8 ==
You can loop through them using a shell script:
for i in `ls -lt | awk '{if ($8 == 2006) print $9}'`;do rm $i;done
-Derek
At 06:35 PM 3/5/2007, Gary Kline wrote:
Guys,
Having found $9 , how do I /bin/rm it (using system()--yes??)
in an awk one-liner?
Gary Kline wrote:
Guys,
Having found $9 , how do I /bin/rm it (using system()--yes??)
in an awk one-liner?
I'm trying to remove from packages from long ago and find and
print them with
ls -lt | awk '{if ($8 == 2006) print $9}';
but what
On Tue, Mar 06, 2007 at 07:27:56AM -0600, Derek Ragona wrote:
You can loop through them using a shell script:
for i in `ls -lt | awk '{if ($8 == 2006) print $9}'`;do rm $i;done
This is the safest way to rm or rm -i each file ($i); the
ls -ls | [awkstuff] spits out the entire
On Mon, Mar 05, 2007, Gary Kline wrote:
Guys,
Having found $9 , how do I /bin/rm it (using system()--yes??)
in an awk one-liner?
I'm trying to remove from packages from long ago and find and
print them with
ls -lt | awk '{if ($8 == 2006) print $9}';
On Mar 5, 2007, at 4:35 PM, Gary Kline wrote:
Having found $9 , how do I /bin/rm it (using system()--yes??)
in an awk one-liner?
I gather that you are looking under /var/db/pkg...?
I'm trying to remove from packages from long ago and find and
print them with
On Mon, Mar 05, 2007 at 04:46:35PM -0800, Chuck Swiger wrote:
On Mar 5, 2007, at 4:35 PM, Gary Kline wrote:
Having found $9 , how do I /bin/rm it (using system()--yes??)
in an awk one-liner?
I gather that you are looking under /var/db/pkg...?
I'm trying to remove from packages
On Mar 6, 2006, at 4:45 PM, Noel Jones wrote:
On 3/6/06, Bart Silverstrim [EMAIL PROTECTED] wrote:
I'm totally drawing a blank on where to start out on this.
If I have a list of URLs like
http://www.happymountain.com/archive/digest.gif
How could I use Awk or Sed to strip everything after
Bart Silverstrim wrote:
On Mar 6, 2006, at 4:45 PM, Noel Jones wrote:
On 3/6/06, Bart Silverstrim [EMAIL PROTECTED] wrote:
I'm totally drawing a blank on where to start out on this.
If I have a list of URLs like
http://www.happymountain.com/archive/digest.gif
How could I use Awk or Sed
On 3/6/06, Bart Silverstrim [EMAIL PROTECTED] wrote:
I'm totally drawing a blank on where to start out on this.
If I have a list of URLs like
http://www.happymountain.com/archive/digest.gif
How could I use Awk or Sed to strip everything after the .com? Or is
there a better way to do it?
Hi Alexandre:
On Friday 20 January 2006 16:59, Alexandre Vieira wrote:
Hello folks,
I'm making a script to generate some statistics for a batch job and I'm
stuck with awk.
For example:
%echo 1 2 3 4 5 6 | awk {'print $1 $2 $3 $4 $5 $6'}
it will output:
1 2 3 4 5 6
I want to tokenize
On 1/20/06, Don Hinton [EMAIL PROTECTED] wrote:
Hi Alexandre:
On Friday 20 January 2006 16:59, Alexandre Vieira wrote:
Hello folks,
I'm making a script to generate some statistics for a batch job and I'm
stuck with awk.
For example:
%echo 1 2 3 4 5 6 | awk {'print $1 $2 $3 $4
# [EMAIL PROTECTED] / 2003-12-15 16:30:33 -0700:
i would like to do something like
df | awk '{print $1}'
to capture all the current file systems. But I would like to strip
off the first and last lines, since these are generally -- not needed.
the goal is to write a generalized script
man head
and
man tail
At 05:30 PM 12/15/2003, you wrote:
i would like to do something like
df | awk '{print $1}'
to capture all the current file systems. But I would like to strip
off the first and last lines, since these are generally -- not needed.
the goal is to write a generalized script
On Mon, 15 Dec 2003, David Bear wrote:
i would like to do something like
df | awk '{print $1}'
to capture all the current file systems. But I would like to strip
off the first and last lines, since these are generally -- not needed.
df | awk '$1 ~/^\/dev/ {print $1}'
Fer
On Mon, Dec 15, 2003 at 08:39:06PM -0300, Fernando Gleiser wrote:
On Mon, 15 Dec 2003, David Bear wrote:
i would like to do something like
df | awk '{print $1}'
to capture all the current file systems. But I would like to strip
off the first and last lines, since these are
37 matches
Mail list logo
