print nth line after the line which matches the string

cristalp · 10-26-2011, 02:11 PM

Dear Experts,

I have a file which looks like:

Code:

What I want is to print the third line (which should be 666666) after the line where aaa was found. I mean, I need to first search through the file to find a line includes string "aaa", then print out the third (or nth) line 666666 after this line.

I tired:

Code:

awk '{a[NR] = $0} $0 ~ /aaa/ {printf("%s\n", a[NR+3])}' FILENAME

This gave me empty output. But if I try for example:

Code:

awk '{a[NR] = $0} $0 ~ /aaa/ {printf("%s\n", a[NR-1])}' FILENAME

With this command, I can get the line 222222 which is above the line with aaa. (exactly NR-1).

So, why I can not do this same way for line after(NR+n)? How could I achieve my goal, with awk or grep(I am not that in sed)?

Any reply would be greatly appreciated. Thank you very much!

druuna · 10-26-2011, 02:51 PM

Hi,

Try this:

Code:

grep -A3 "aaa" infile | tail -1

Hope this helps.

grail · 10-27-2011, 12:28 AM

The issue with your awks is you can see in the past (ie what you have stored in the array), but you cannot see into the future (ie NR +3 has yet to be stored).

What you would try and do if you use awk is:

Code:

awk '/aaa/{x = NR + 3}NR == x' file

Nominal Animal · 10-27-2011, 01:37 AM

I'm with grail, except that if you have multiple matches per file, I'd use an indicator array:

Code:

awk '/aaa/ { show[NR+3] = 1 } (NR in show) { print $0 ; delete show[NR] }' infile

The delete will trim the array, so it'll ever only contain the to-be-printed record numbers. It's a good idea if you happen to process files with many occurrences, but basically it's just an optimization.

If you have multiple trigger patterns, you could even do

Code:

awk '/aaa/ { show[NR+3] = "aaa" }
     /bbb/ { show[NR+3] = "bbb" }
     (NR in show) { printf("%s: %s\n", show[NR], $0) ; delete show[NR] }
    ' infile

although in this simple form it wont tell you if a line was printed for more than one trigger. To list all reasons, you'd need to use for example

Code:

awk '/aaa/ { if ((NR+3) in show) show[NR+3] = show[NR+3] ", aaa" ; else show[NR+3] = "aaa" }
     /bbb/ { if ((NR+3) in show) show[NR+3] = show[NR+3] ", bbb" ; else show[NR+3] = "bbb" }
     (NR in show) { printf("%s: %s\n", show[NR], $0) ; delete show[NR] }
    ' infile

expertshell · 10-27-2011, 02:34 AM

grail's is a good solution

grail · 10-27-2011, 07:00 AM

Yes but where mine gets caught is if the pattern appears more than once prior to the line required:

Code:

Here you should get the 6s and the 7s but mine will only print the last entry. A slight alteration you can make to NA's is:

Code:

awk '/aaa/ { show[NR+3]++  } show[NR]' file

cristalp · 10-27-2011, 01:19 PM

Thanks everyone! Your kind and excellent answers expand this small question to a big tank of knowledge and skills. I am glad to see that we share our knowledge so broadly. This is why I join this forum and I hope every one got her progress through the communication.

I, myself, also found another way to solve this problem from a similar problem in another forum:

Code:

grep -A 3 aaa FILENAME | sed -n '4~5p'

Anybody know the meaning of '4~5p'?

Let's keep this question unsolved for a bit while, since I still have some questions on your replies. I will put my questions later on. Please wait for a bit...

David the H. · 10-27-2011, 01:53 PM

First of all, useless use of grep, since sed can also match text.

Second, did you look in the sed man page? It explains what that pattern means.

But I'd use A1,+N instead, since you want a fixed number of lines after the match, rather than a step.

So use one address range to grab that section of lines, then apply a second nested command to it to print just the line you want.

Code:

sed -n '/aaa/,+3 { 1,+2d ; p }' file.txt

I don't know why, but the nested address matching is kind of tricky. I can't get it to just directly print the last line of the initial match ($ still seems to mean the last line of the file), so you have to tell it to delete everything before it instead.