Re: awk question: replacing %d%s by %d %s
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 get more and more stupid myself for NOT being able to solve this, even after... some nearly infinite time. :-) I have strings of the form either number(s) or number(s)letter. I catch them with if(match(nr, [a-z])) ... 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 would get 6 d, and 58 would stay unchanged. I've tried with split(), with array manipulation and could produce 10 lines of code that didn't work as intended (it produced 1122aa, 66dd and 5588 according to the examples above). Obviously, sub(nr, [a-z], [a-z]); is nonsense. 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 pattern regex'. ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to freebsd-questions-unsubscr...@freebsd.org
Re: awk question: replacing %d%s by %d %s
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 pattern regex'. Doesn't work on my 7-STABLE system (20080811), awk stops: awk: syntax error at source line 78 source file konvertieren.awk context is sub(nr, [a-z], ); awk: illegal statement at source line 79 source file konvertieren.awk But I'll keep your suggestion in the program source and test it on 8 as soon as my new home system is ready to use. At least, the variant looks much better. -- Polytropon Magdeburg, Germany Happy FreeBSD user since 4.0 Andra moi ennepe, Mousa, ... ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to freebsd-questions-unsubscr...@freebsd.org
Re: awk question: replacing %d%s by %d %s
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 regex library that means 'that which was matched by the pattern regex'. Doesn't work on my 7-STABLE system (20080811), awk stops: awk: syntax error at source line 78 source file konvertieren.awk context is sub(nr, [a-z], ); awk: illegal statement at source line 79 source file konvertieren.awk But I'll keep your suggestion in the program source and test it on 8 as soon as my new home system is ready to use. At least, the variant looks much better. 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 is not given, $0 is used. So the correct syntax is: sub([a-z], , nr) Wayne ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to freebsd-questions-unsubscr...@freebsd.org
Re: awk question: replacing %d%s by %d %s
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 is not given, $0 is used. So the correct syntax is: sub([a-z], , nr) Works in this version, thanks! Reduction of 7 lines of code. sub(from, to, where); is the correct form. -- Polytropon Magdeburg, Germany Happy FreeBSD user since 4.0 Andra moi ennepe, Mousa, ... ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to freebsd-questions-unsubscr...@freebsd.org
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 get more and more stupid myself for NOT being able to solve this, even after... some nearly infinite time. :-) I have strings of the form either number(s) or number(s)letter. I catch them with if(match(nr, [a-z])) ... 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 would get 6 d, and 58 would stay unchanged. I've tried with split(), with array manipulation and could produce 10 lines of code that didn't work as intended (it produced 1122aa, 66dd and 5588 according to the examples above). Obviously, sub(nr, [a-z], [a-z]); is nonsense. Can anybody please tell me the obvious and easy way which I am missing? Thanks a lot! =^_^= The reason I'd like to do this in awk is that I've already spent enough time creating a format converter in awk that works very fine, as intended, but is missing just this little feature. -- Polytropon Magdeburg, Germany Happy FreeBSD user since 4.0 Andra moi ennepe, Mousa, ... ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to freebsd-questions-unsubscr...@freebsd.org
Re: awk question: replacing %d%s by %d %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 lines. If that
is bad, try:
$ awk data.txt experiment.txt '{ num = $1 ; sub(/[^0-9]+$/, ,
num) ; lets = $1 ; sub(/^[0-9]+/, , lets); if (length(lets)) { print
num lets } else { print num } }' ; diff control.txt experiment.txt
$
Thanks for your inspiration! I concluded that it is
a possible way to split the string into two parts:
not-numbers (letters) and numbers, and then recombining
them in the desired way.
Here's my code:
if(match(nr, [a-z])) {
z = nr;
gsub(/[^0-9]+$/, , z);
b = nr;
gsub(/[0-9]/, , b);
nr = sprintf(%s %s, z, b);
}
First catch all nr's that have a letter in it, then
remove the letters in one copy, then the numbers in
another one, and finally overwrite nr with the proper
recombination that includes the space.
Works as intended, FULLY. Thanks!
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Magdeburg, Germany
Happy FreeBSD user since 4.0
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Re: awk question: replacing %d%s by %d %s
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 would
get 6 d, and 58 would stay unchanged. I've tried
This feels like it could be faster, but is reasonable:
$ cat data.txt
1
12
3
1d
1dc
12d
12dc
123d
123dc
123dcb
$ cat control.txt
1
12
3
1 d
1 dc
12 d
12 dc
123 d
123 dc
123 dcb
$ 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 lines. If that
is bad, try:
$ awk data.txt experiment.txt '{ num = $1 ; sub(/[^0-9]+$/, ,
num) ; lets = $1 ; sub(/^[0-9]+/, , lets); if (length(lets)) { print
num lets } else { print num } }' ; diff control.txt experiment.txt
$
Tom
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Re: awk question (actively tail a file notify when expression is found)
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 /var/log/apache2/access.log | awk '/192.168.1.100/ {
system(sprintf(echo %s | mail m...@email.address, $0)); }'
Any pointers would be much appreciated.
It should work, but I'm sure someonle else will soon show you
some more elegant way. :-)
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awk question (actively tail a file notify when expression is found)
Greetings..!
this works,
tail -f /var/log/apache2/access.log | nawk '/192.168.1.100/{ print $0 }'
and this too:
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { system
(mail -s \This works\ m...@email.address)}'
but this below, does not work
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { print
$0 | mail m...@email.address }'
Any pointers would be much appreciated.
Thx.
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Re: awk question (actively tail a file notify when expression is found)
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:
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { system
(mail -s \This works\ m...@email.address)}'
but this below, does not work
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { print
$0 | mail m...@email.address }'
Any pointers would be much appreciated.
Thx.
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Re: awk question (actively tail a file notify when expression is found)
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 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:
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { system
(mail -s \This works\ m...@email.address)}'
but this below, does not work
tail -f /var/log/apache2/access.log |awk ' /192.168.1.100/ { print
$0 | mail m...@email.address }'
Any pointers would be much appreciated.
Thx.
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Voice: (206) 236-1676 Mercer Island, WA 98040-0820
Fax: (206) 232-9186 Skype: jwccsllc (206) 855-5792
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scum end up being extremely bad for business when there's working
competition. -rra
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awk question
Len Conrad lcon...@go2france.com wrote:
We need to print a line when the 3rd field (with trailing ;
delimiter) is, eg, exactly 5 lower case characters
awk ' $3 ~ /^[a-z]{5,5};$/ {print $0} ' file
... doesn't work.
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
If that still doesn't work for you, please specify your
file format more exactly, and provide an example of the
input lines.
Best regards
Oliver
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Re: awk question
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 default action
(to print the whole line, i. e. $0) when no action is given
for a pattern.
% awk -F';' '$3 ~ /^[a-z]{5}$/' file
When using this in a shell, keep an eye on eventually needed
quoting or escaping of $.
--
Polytropon
From Magdeburg, Germany
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Andra moi ennepe, Mousa, ...
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Re: awk question
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 http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to [EMAIL PROTECTED]
Re: awk question
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 $(NF-1) }' user.csv
--
Nino
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Re: awk question
n j writes:
Or awk only i.e. no sed:
awk '!(/^$/) { 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
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Re: awk question
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
source line number 1
Obviously $(NF-1) doesn't do the trick. Any better idea?
Hmmm, works for me it does...
Peter
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Re: awk question
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 out of range field -1
input record number 1, file user.csv
source line number 1
Obviously $(NF-1) doesn't do the trick. Any better idea?
Hmmm, works for me it does...
Me too, except of course if the first line of user.cvs is blank...
Peter
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Re: awk question
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
input record number 1, file user.csv
source line number 1
Obviously $(NF-1) doesn't do the trick. Any better idea?
Hmmm, works for me it does...
Sorry: Actually the first line of my user.csv was empty.
Thanks though,
Uli.
Peter
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Peter Ulrich Kruppa
Wuppertal
Germany
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Re: awk question
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 WB5AGZ Stillwater, OK
Systems Engineer
OSU Information Technology Department Network Operations Group
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awk question
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
source line number 1
Obviously $(NF-1) doesn't do the trick. Any better idea?
Thanks,
Uli.
Peter Ulrich Kruppa
Wuppertal
Germany
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Re: awk question
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 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 file in my ~/HowTo,
thankee much. I'm in the first stages on a months-long trial on
system tuning. This, before I'd risk publishing anything. So
far tho, by upping and lower the NICE prio of various binaries, I
have been able to get more than 70% efficient use out of my older
servers. ---This *ought* to carry over to my faster machines
Is tthere a way of using ps -alx | ask to look at nice and if it
is non-zero (the default), to reset it to zero?
You can easily do some of this using top, such as:
top -bS 200 | tail -n +9 | awk '{ print $5 }'
If you want to tweak the nice value you'd need to examine the value and
then renice it as long as you are root. You'd need the PID for that, so
here's another example:
top -bS 200 | tail -n +9 | awk '{ printf(Pid: %d has Nice: %d\n,
$1,$5) }'
Well, I knew there had to be a static way to read top. -bS is
it. If NICE is 9, then renice-n -9 pid ought to reset it to 0;
so in C, the check for nice or n would be trivial:
if (n != 0)
n = -n;
In you example, would this be if ($1 != 0) $1 = -$1;
then a '{system(renice -n $)};
or is this disallowed in awk?
gary
It is easier to redirect the output to a file then just execute that
file. You'd usually have this in a shell script run by cron.
top -bS 200 | tail -n +9 | awk '{ if ($1 != 0) printf(/usr/bin/renice %d
%d\n, $1,-$5) }' /tmp/renice.scr
sh -c /tmp/renice.scr
But look at the file generated, you need to do more than just the check for
0 and then negate it.
-Derek
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Re: awk 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 file in my ~/HowTo,
thankee much. I'm in the first stages on a months-long trial on
system tuning. This, before I'd risk publishing anything. So
far tho, by upping and lower the NICE prio of various binaries, I
have been able to get more than 70% efficient use out of my older
servers. ---This *ought* to carry over to my faster machines
Is tthere a way of using ps -alx | ask to look at nice and if it
is non-zero (the default), to reset it to zero?
Say that I've reniced xorg and mozilla to -9 for starters and
reniced epiphany to -11. Others may have a much lower NICE of
19. I don't know ask that well; is this on better left to
something like C? :-) [[[ Something I can do?? :-) ]]]
thanks in advance,
gary
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Re: awk question
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 save your snippet to my growing %%% awk file in my ~/HowTo,
thankee much. I'm in the first stages on a months-long trial on
system tuning. This, before I'd risk publishing anything. So
far tho, by upping and lower the NICE prio of various binaries, I
have been able to get more than 70% efficient use out of my older
servers. ---This *ought* to carry over to my faster machines
Is tthere a way of using ps -alx | ask to look at nice and if it
is non-zero (the default), to reset it to zero?
You can easily do some of this using top, such as:
top -bS 200 | tail -n +9 | awk '{ print $5 }'
If you want to tweak the nice value you'd need to examine the value and
then renice it as long as you are root. You'd need the PID for that, so
here's another example:
top -bS 200 | tail -n +9 | awk '{ printf(Pid: %d has Nice: %d\n, $1,$5) }'
-Derek
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Re: awk question
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 following:
ls -l | awk '$8 == 2006 {system(rm $9)}'
i'Ll save your snippet to my growing %%% awk file in my ~/HowTo,
thankee much. I'm in the first stages on a months-long trial on
system tuning. This, before I'd risk publishing anything. So
far tho, by upping and lower the NICE prio of various binaries, I
have been able to get more than 70% efficient use out of my older
servers. ---This *ought* to carry over to my faster machines
Is tthere a way of using ps -alx | ask to look at nice and if it
is non-zero (the default), to reset it to zero?
You can easily do some of this using top, such as:
top -bS 200 | tail -n +9 | awk '{ print $5 }'
If you want to tweak the nice value you'd need to examine the value and
then renice it as long as you are root. You'd need the PID for that, so
here's another example:
top -bS 200 | tail -n +9 | awk '{ printf(Pid: %d has Nice: %d\n, $1,$5) }'
Well, I knew there had to be a static way to read top. -bS is
it. If NICE is 9, then renice-n -9 pid ought to reset it to 0;
so in C, the check for nice or n would be trivial:
if (n != 0)
n = -n;
In you example, would this be if ($1 != 0) $1 = -$1;
then a '{system(renice -n $)};
or is this disallowed in awk?
gary
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Re: awk question
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 == 2006) print $9}' )
If this exceeds the maximum length of a line, just use xargs also.
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Re: awk question
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?
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 I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
thanks in advance,
gary
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Re: awk question
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 I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
thanks in advance,
gary
Another way is:
ls -lt | awk '{if ($8 == 2006) print rm -rf $9}' | sh
but I agree, using pkg_delete would be safer:
ls -lt | awk '{if ($8 == 2006) print pkg_delete $9}' | sh
-SO
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Re: awk question
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 list in one chunk.
But since packages from 2006 were OLD, I just /bin/rm'd them
wholesale. Thanks,
gary
-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?
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 I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
thanks in advance,
gary
--
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awk question
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 I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
thanks in advance,
gary
--
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Re: awk question
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}';
but what I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
A better way to do this might be to use find and xargs. The
command below would remove all files under the current directory
that haven't been modified in the 360 days.
find . -type f -mtime +360 -print0 | xargs -0 rm
If you don't want it to go into subdirectories:
find . -maxdepth 1 -type f -mtime +360 -print0 | xargs -0 rm
Bill
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Re: awk question
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
ls -lt | awk '{if ($8 == 2006) print $9}';
but what I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
You could pipe the output of awk through | xargs rm -rf...but be
careful.
Putting it through pkg_delete (-f) might be safer.
--
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Re: awk question
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 from long ago and find and
print them with
ls -lt | awk '{if ($8 == 2006) print $9}';
but what I want to remove the file pointed at by $9. I've tried
FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
What's the magic here?
You could pipe the output of awk through | xargs rm -rf...but be
careful.
Putting it through pkg_delete (-f) might be safer.
Bill and Chuck:
These are a slew of packages from 2006. (Obv'ly in
/usr/ports/packages. I plumb fergot about xargs (again);
just rarely use it.
ls -lt | awk Foo catches those that are almost certainly invalid,
from '06) but the find -mtime [days] is another way. Like they
say about perl, there's always more than one way... When you've
got Unix, you've got power at your fingertips.
thanks, gents,
gary
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Re: awk question
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 the .com? Or is there a better way to do it? I'd like to just pipe the information from the logs to this mini-script and end up with a list of URLs consisting of just the domain (http://www.happymountain.com). | cut -d / -f 1-3 Oh boy was that one easy. It was a BAD mental hiccup. I'll add a sort and uniq and it should be all ready to go. Thanks! ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to [EMAIL PROTECTED]
Re: awk question
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 to strip everything after the .com? Or is
there a better way to do it?
| cut -d / -f 1-3
Oh boy was that one easy. It was a BAD mental hiccup.
I'll add a sort and uniq and it should be all ready to go. Thanks!
More than one way to skin that cat! cut is nice'n'easy but since you
asked about awk and sed, these would work too:
awk -F/ 'NF 2 {printf %s//%s\n, $1, $3}'
or
sed 's,^\([^/]*://[^/]*\).*,\1,'
--Alex
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awk question
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? I'd like to just pipe the information from the logs to this mini-script and end up with a list of URLs consisting of just the domain (http://www.happymountain.com). Any suggestions? ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to [EMAIL PROTECTED]
Re: awk question
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? I'd like to just pipe the information from the logs to this mini-script and end up with a list of URLs consisting of just the domain (http://www.happymountain.com). | cut -d / -f 1-3 -- Noel Jones ___ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to [EMAIL PROTECTED]
awk question
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 a string with another separating char (the : char):
%echo 1:2:3:4:5:6
and with awk to output:
1 2 3 4 5 6
Is there any way of doing this?
Real example:
I have a log file with the following output:
2006-01-20 - 20:01:07 - Some text
2006-01-20 - 20:01:15 - Some text
2006-01-20 - 20:01:38 - Some text
(...)
and since I'm generating hourly stats I need to match the 20 which is in a
string 20:01:07 that is separated by the char :.
I hope I've been clear.
Thanks for your help.
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Re: awk question
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 a string with another separating char (the : char):
%echo 1:2:3:4:5:6
and with awk to output:
1 2 3 4 5 6
Is there any way of doing this?
Sure. Here's a link to the online awk documentation section on field
seperators...
http://www.gnu.org/software/gawk/manual/html_node/Field-Separators.html#Field-Separators
ciao...
don
Real example:
I have a log file with the following output:
2006-01-20 - 20:01:07 - Some text
2006-01-20 - 20:01:15 - Some text
2006-01-20 - 20:01:38 - Some text
(...)
and since I'm generating hourly stats I need to match the 20 which is in
a string 20:01:07 that is separated by the char :.
I hope I've been clear.
Thanks for your help.
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Re: awk question
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 $5 $6'}
it will output:
1 2 3 4 5 6
I want to tokenize a string with another separating char (the : char):
%echo 1:2:3:4:5:6
and with awk to output:
1 2 3 4 5 6
Is there any way of doing this?
Sure. Here's a link to the online awk documentation section on field
seperators...
http://www.gnu.org/software/gawk/manual/html_node/Field-Separators.html#Field-Separators
ciao...
don
Real example:
I have a log file with the following output:
2006-01-20 - 20:01:07 - Some text
2006-01-20 - 20:01:15 - Some text
2006-01-20 - 20:01:38 - Some text
(...)
and since I'm generating hourly stats I need to match the 20 which is
in
a string 20:01:07 that is separated by the char :.
I hope I've been clear.
Thanks for your help.
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Geez I can believe it was that easy. Thank you.
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An AWK question
--- FBSD-4.10p2
I have a script that tells me when a mailbox exceeds 2MB. Using it for
another purpose. how can I modify this script to tell me when a file is
LESS than a certain size?
-
#!/bin/sh
#AWK=/usr/bin/awk
#FILE=/file/size/to/check
#LS=/bin/ls
#SIZE=2048 # (2048 = 2MB)
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -gt ${SIZE} ]; then
#echo File ${FILE} is ABOVE normal | mail -s Alert: Check the file
#fi
-
Thanks, and
Best regards,
Jack L. Stone
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Re: An AWK question
On Fri, Feb 04, 2005 at 01:14:57PM -0600, antenneX wrote:
--- FBSD-4.10p2
I have a script that tells me when a mailbox exceeds 2MB. Using it for
another purpose. how can I modify this script to tell me when a file is
LESS than a certain size?
-
#!/bin/sh
#AWK=/usr/bin/awk
#FILE=/file/size/to/check
#LS=/bin/ls
#SIZE=2048 # (2048 = 2MB)
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -gt ${SIZE} ]; then
#echo File ${FILE} is ABOVE normal | mail -s Alert: Check the file
#fi
-
Just change the word -gt to -lt.
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Re: An AWK question
On Friday 04 February 2005 02:14 pm, antenneX wrote:
--- FBSD-4.10p2
I have a script that tells me when a mailbox exceeds 2MB. Using it for
another purpose. how can I modify this script to tell me when a file is
LESS than a certain size?
-
#!/bin/sh
#AWK=/usr/bin/awk
#FILE=/file/size/to/check
#LS=/bin/ls
#SIZE=2048 # (2048 = 2MB)
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -gt ${SIZE} ]; then
#echo File ${FILE} is ABOVE normal | mail -s Alert: Check the file
#fi
-
Thanks, and
Best regards,
Jack L. Stone
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Jonathen's correct. It's not an awk question. It's actually being evaluated by
test. see man test )test is also invoked when you see [ some equation ] (you
need the space at least on the inside of [ and ], perhaps on the outsides
too.
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -lt ${SIZE} ]; then
#echo File ${FILE} is BELOW normal | mail -s Alert: Check the file
#fi
awk is just being used to get the first field from ls -s, which is the size...
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Re: An AWK question
- Original Message -
From: Steven Friedrich [EMAIL PROTECTED]
To: freebsd-questions@freebsd.org; antenneX [EMAIL PROTECTED]
Sent: Friday, February 04, 2005 2:23 PM
Subject: Re: An AWK question
On Friday 04 February 2005 02:14 pm, antenneX wrote:
--- FBSD-4.10p2
I have a script that tells me when a mailbox exceeds 2MB. Using it
for
another purpose. how can I modify this script to tell me when a file
is
LESS than a certain size?
-
#!/bin/sh
#AWK=/usr/bin/awk
#FILE=/file/size/to/check
#LS=/bin/ls
#SIZE=2048 # (2048 = 2MB)
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -gt ${SIZE} ]; then
#echo File ${FILE} is ABOVE normal | mail -s Alert: Check the
file
#fi
-
Thanks, and
Best regards,
Jack L. Stone
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Jonathen's correct. It's not an awk question. It's actually being
evaluated by
test. see man test )test is also invoked when you see [ some
equation ] (you
need the space at least on the inside of [ and ], perhaps on the
outsides
too.
#if [ `${LS} -s ${FILE} | ${AWK} '{ print $1 }'` -lt ${SIZE} ]; then
#echo File ${FILE} is BELOW normal | mail -s Alert: Check the
file
#fi
awk is just being used to get the first field from ls -s, which is the
size...
--
No wonder I couldn't make sense out of it looking only in man GAWK(1).
Now, it's very simple -- thanks a lot!
Jack
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Re: awk question, maybe
# [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 that can feed dump with all
the file systems on a box.
I would use mount(8) instead: its output doesn't include any
headers, but you'll still want to pass through only local
filesystems, so it doesn't really matter:
[EMAIL PROTECTED] ~ 1006:0 mount | awk '/\/dev\/(ad|da)/ {print $1}'
/dev/ad0s1a
/dev/ad0s1e
[EMAIL PROTECTED] ~ 1007:0 df | awk '/\/dev\/(ad|da)/ {print $1}'
/dev/ad0s1a
/dev/ad0s1e
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Re: awk question, maybe
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 that can feed dump with all
the file systems on a box.
any pointers?
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fax:480-965-9189
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awk question, maybe
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 that can feed dump with all
the file systems on a box.
any pointers?
--
David Bear
phone: 480-965-8257
fax:480-965-9189
College of Public Programs/ASU
Wilson Hall 232
Tempe, AZ 85287-0803
Beware the IP portfolio, everyone will be suspect of trespassing
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Re: awk question, maybe
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
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Re: awk question, maybe
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 generally -- not needed.
df | awk '$1 ~/^\/dev/ {print $1}'
yes.. I was hoping it would be simple. and I thought of using a regex
to match the lines I want after sending that note.
thanks for all the replies.
--
David Bear
phone: 480-965-8257
fax:480-965-9189
College of Public Programs/ASU
Wilson Hall 232
Tempe, AZ 85287-0803
Beware the IP portfolio, everyone will be suspect of trespassing
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