SED: addressing two lines before match

Question

Print line, which is situated 2 lines before the match(pattern).

I tried next:

sed -n \': loop
/.*/h
:x
{n;n;/cen/p;}
s/./c/p
t x
s/n/c/p
t loop
{g;p         


        

Answer 1

Here are some other variants:

awk '{a[NR]=$0} /pattern/ {f=NR} END {print a[f-2]}' file

This stores all lines in an array a. When pattern is found store line number.
At then end print that line number from the file.
PS may be slow with large files

---

Here is another one:

awk 'FNR==NR && /pattern/ {f=NR;next} f-2==FNR' file{,}

This reads the file twice (file{,} is the same as file file)
At first round it finds the pattern and store line number in variable f
Then at second round it prints the line two before the value in f