Hello ,

I need your help to extract a line in a big file , and this line is always 11 lines
before a specific pattern . Do you know a way via Awk ?

Thanks in advance

npn35

mate try this,

awk 'BEGIN {}^J{ if( NR == 11 ) print $0 };^JEND{} ' datafile

Cheers,

In fact , I don't want to print the line 11 . I need to search for a specific pattern
then print the line NR-11 .
I hope my explanation is better .

Thanks.
BR

Try this...

/pattern/ && NR == 11 {
print $0;
}

Thanks but the pattern is not located on line 11 . I want to print the line
which is 11 lines before the pattern .

Regards
npn35

hey man u r not so clearabout what u want?cleat itmate.then it would be easier to give u thesolution.

I want this

I a file with lines containing 3 columns each,

If column 3 of a line is zero, then
substract the second entry of present column from
the second entry of the prvious column.

e.g.

28 -4.8743 2
29 -4.8726 2
30 -4.7833 2
31 -4.7721 2
32 -4.7690 2
33 -4.6266 2
34 -4.6203 2
35 -3.8990 1
36 -3.2009 0
37 -2.9944 0
38 -2.9934 0
39 -2.8856 0
40 -2.8820 0

then it should subtract -3.2009 from -3.8990 and give me the answer.