[SOLVED] AWK: print to end staring from nth line after matching

cristalp · 08-24-2012, 10:31 AM

Dear Experts,

I have a file like:

Code:

ssssss
oooooo
rrrrrr
aaa
111111
222222
333333
444444
555555
666666
END

What I want is to locate the pattern "aaa" then from there I skip n lines and print remaining lines to the end of the file.

Say, I'd like to skip 3 lines after matching "aaa", then I would expect a result like

Code:

I tried the code:

Code:

awk '/aaa/{n=NR} NR > n+3 {print}' INPUTFILE

But it printed me all the lines in the file. What was the wrong with this code? And how could I achieve what I want in a simple way with AWK?

I would thank you for your help!

byannoni · 08-24-2012, 10:37 AM

Code:

awk -v i=0 '/aaa/ {i=1} i==1' INPUTFILE

firstfire · 08-24-2012, 10:40 AM

Hi.

Your awk seems to work well:

Code:

$ cat in
ssssss
oooooo
rrrrrr
aaa
111111
222222
333333
444444
555555
666666
END
$ awk '/aaa/{n=NR} NR > n+3 {print}' in
444444
555555
666666
END

BTW, you can omit {print} safely:

Code:

$ awk '/aaa/{n=NR} NR > n+3' in

because matched lines are printed by default.

EDIT: I got it.. n remains undefined (thus zero) until awk encounters /aaa/, so if you have more than 3 lines before 'aaa', they will be printed. Workaround:

Code:

$ cat in
ssssss
ssssss
ssssss
ssssss
oooooo
rrrrrr
aaa
111111
222222
333333
444444
555555
666666
END
$ awk '/aaa/{n=NR;} NR > n+3' n=1e5 in
444444
555555
666666
END

Here we initialize `n' to be large enough. Other, IMO more correct way is to check if n is not zero:

Code:

$ awk '/aaa/{n=NR;} n && NR > n+3' in

David the H. · 08-24-2012, 04:08 PM

For the record, here's a sed solution as well.

Code:

sed -n '/aaa/,$ {/aaa/,+3 d;p}' infile

It first matches from line aaa to the end, then from that selection it deletes the aaa line and the three following it, and prints the remainder.

Note that the +" address option is, I believe, a gnu sed extension, and other implementations likely don't have it.

And to round it off, here are two ways to do it in ed too.

Code:

printf '%s\n' '1,/aaa/+3d' '%p' | ed -s infile
printf '%s\n' '/aaa/+4,$p' | ed -s infile

The first one deletes everything from line 1 to aaa+3, and prints what remains, while the second simply prints from line aaa+4 to the end (the rest of the lines are still in the buffer). As you can see, ed has a more flexible address system than sed.

If you replace the '%p/p' commands with 'w', you can save the changes directly back to the input file. Or use 'w outfile' to write them to a new file. The first version is probably slightly better for this.

How to use ed:
http://wiki.bash-hackers.org/howto/edit-ed
http://snap.nlc.dcccd.edu/learn/nlc/ed.html
(also read the info page)

cristalp · 08-28-2012, 04:51 AM

Quote:

Originally Posted by firstfire

Hi.

Your awk seems to work well:

Code:

$ cat in
ssssss
oooooo
rrrrrr
aaa
111111
222222
333333
444444
555555
666666
END
$ awk '/aaa/{n=NR} NR > n+3 {print}' in
444444
555555
666666
END

BTW, you can omit {print} safely:

Code:

$ awk '/aaa/{n=NR} NR > n+3' in

because matched lines are printed by default.

EDIT: I got it.. n remains undefined (thus zero) until awk encounters /aaa/, so if you have more than 3 lines before 'aaa', they will be printed. Workaround:

Code:

$ cat in
ssssss
ssssss
ssssss
ssssss
oooooo
rrrrrr
aaa
111111
222222
333333
444444
555555
666666
END
$ awk '/aaa/{n=NR;} NR > n+3' n=1e5 in
444444
555555
666666
END

Here we initialize `n' to be large enough. Other, IMO more correct way is to check if n is not zero:

Code:

$ awk '/aaa/{n=NR;} n && NR > n+3' in

Thanks!! This fully explains my question! Very clear. No confusion any more.