How to display a file, omitting lines that contain a string?

Thelionroars · 11-17-2009, 09:52 PM

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(#).

ghostdog74 · 11-17-2009, 09:59 PM

awk, grep are 2 of them. read their docs and try to code them yourself first.

Thelionroars · 11-17-2009, 10:20 PM

<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?

ghostdog74 · 11-17-2009, 10:51 PM

Quote:

Originally Posted by Thelionroars

So:

Code:

cat /boot/grub/menu.lst | grep -v '#'

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?

yes.

Thelionroars · 11-18-2009, 05:53 PM

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.

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.

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?

smeezekitty · 11-18-2009, 05:56 PM

you can make it pause so you an actually read it by replacing cat with more. just a tip.

ghostdog74 · 11-18-2009, 06:43 PM

i would use gawk instead

Code:

awk '!/^ +#/' file  # one or spaces in beginning, don't print

smeezekitty · 11-18-2009, 06:51 PM

Quote:

Originally Posted by ghostdog74

i would use gawk instead

Code:

awk '!/^ +#/' file  # one or spaces in beginning, don't print

the grep version looked cleaner.

ghostdog74 · 11-18-2009, 06:58 PM

Quote:

Originally Posted by smeezekitty

the grep version looked cleaner.

and so? awk can do a whole lot more.

Thelionroars · 11-18-2009, 08:42 PM

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.

Fabio Paolini · 01-21-2010, 08:11 AM

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.

chrism01 · 01-21-2010, 05:42 PM

Actually, '|' is the 'OR' operator

Fabio Paolini · 01-22-2010, 07:03 AM

Quote:

Originally Posted by chrism01

Actually, '|' is the 'OR' operator

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