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11-17-2009, 09:52 PM
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#1
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LQ Newbie
Registered: Aug 2009
Location: Melbourne, Australia
Distribution: Ubuntu
Posts: 19
Rep:
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How to display a file, omitting lines that contain a string?
What would be the simplest command/s to display a file, with lines containing a certain string omitted?
For example - I would like to display my GRUB file in a terminal in Ubuntu 9.04 (cat /boot/grub/menu.lst) but don't want to display the comments(#).
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11-17-2009, 09:59 PM
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#2
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Senior Member
Registered: Aug 2006
Posts: 2,697
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awk, grep are 2 of them. read their docs and try to code them yourself first.
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11-17-2009, 10:20 PM
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#3
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LQ Newbie
Registered: Aug 2009
Location: Melbourne, Australia
Distribution: Ubuntu
Posts: 19
Original Poster
Rep:
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<ghostdog74 thrusts fishing rod into thelionroars' hands>
Ok ok... a quick scan of the grep manpage showed me what I wanted (use the -v option). So:
Code:
cat /boot/grub/menu.lst | grep -v '#'
Thanks. Would your username be a reference to the film?
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11-17-2009, 10:51 PM
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#4
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Senior Member
Registered: Aug 2006
Posts: 2,697
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Quote:
Originally Posted by Thelionroars
So:
Code:
cat /boot/grub/menu.lst | grep -v '#'
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good effort. note: no need cat. Its useless. many i have come across always do that. Just pass the file to grep. Next, you may have "#" as comment at the back of the line. if you do -v "#" , then you may omit an actual line.
Quote:
Would your username be a reference to the film?
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yes.
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11-18-2009, 05:53 PM
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#5
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LQ Newbie
Registered: Aug 2009
Location: Melbourne, Australia
Distribution: Ubuntu
Posts: 19
Original Poster
Rep:
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Quote:
Originally Posted by ghostdog74
note: no need cat. Its useless. many i have come across always do that. Just pass the file to grep.
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Maybe its the Ubuntu permissions system, but the command won't work without cat, even with sudo. Without sudo, permission denied. With sudo, command not found. Maybe its different using a shell owned by root?
Quote:
Next, you may have "#" as comment at the back of the line. if you do -v "#" , then you may omit an actual line.
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Good point. So I could use a caret (^) to specify that grep only selects lines that begin with the string -
cat /boot/grub/menu.lst | grep ^#
The material on regular expressions that I'm looking at also states that you can add an OR operator with | , but I'm not sure how to use this with grep?
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11-18-2009, 05:56 PM
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#6
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Senior Member
Registered: Sep 2009
Location: Washington U.S.
Distribution: M$ Windows / Debian / Ubuntu / DSL / many others
Posts: 2,339
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you can make it pause so you an actually read it by replacing cat with more. just a tip.
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11-18-2009, 06:43 PM
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#7
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Senior Member
Registered: Aug 2006
Posts: 2,697
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i would use gawk instead
Code:
awk '!/^ +#/' file # one or spaces in beginning, don't print
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11-18-2009, 06:51 PM
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#8
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Senior Member
Registered: Sep 2009
Location: Washington U.S.
Distribution: M$ Windows / Debian / Ubuntu / DSL / many others
Posts: 2,339
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Quote:
Originally Posted by ghostdog74
i would use gawk instead
Code:
awk '!/^ +#/' file # one or spaces in beginning, don't print
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the grep version looked cleaner.
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11-18-2009, 06:58 PM
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#9
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Senior Member
Registered: Aug 2006
Posts: 2,697
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Quote:
Originally Posted by smeezekitty
the grep version looked cleaner.
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and so? awk can do a whole lot more.
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11-18-2009, 08:42 PM
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#10
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LQ Newbie
Registered: Aug 2009
Location: Melbourne, Australia
Distribution: Ubuntu
Posts: 19
Original Poster
Rep:
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I see what you mean now by not needing cat - you give the file name as an argument to grep or awk and it goes to standard output, which by default is printing to the terminal. So to correct my grep usage -
grep -v ^# /boot/grub/menu.lst
No cat or pipes needed.
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01-21-2010, 08:11 AM
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#11
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Member
Registered: Dec 2008
Location: Brazil
Distribution: Slackware 12 Debian 5
Posts: 52
Rep:
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As I recently learned a little about grep command and regular expressions, I would like to improve a bit the command above.
You can also skip the blank lines with
Code:
grep -E -v '^#|^ *$' <file>
The | character is a kind of 'and' operator and the string ^ *$ matches blank lines. At last, if I am right, the 'and' operator works just if the -E option is included.
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01-21-2010, 05:42 PM
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#12
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LQ Guru
Registered: Aug 2004
Location: Sydney
Distribution: Rocky 9.x
Posts: 18,443
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Actually, '|' is the 'OR' operator

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01-22-2010, 07:03 AM
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#13
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Member
Registered: Dec 2008
Location: Brazil
Distribution: Slackware 12 Debian 5
Posts: 52
Rep:
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Quote:
Originally Posted by chrism01
Actually, '|' is the 'OR' operator

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Yes, that is true, actually I said 'and' without thinking enough.
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