Shell Programming and Scripting

i want to search a log for a string. when that string is found, i want to grab the a set number of lines that came before the string, and a set number of lines that come after the string.

so if i search for the word "Error" in the /var/log/messages file, how can I output the 20 lines that came directly before it and the 35 lines that come after it..and also output the line that contained the specified string of "Error"?

im thinking a mixture of sed and awk can work here?

OS: Linux / SunOS
shell: bash

Code:

$ awk '{a[NR]=$0}/pattern/{for(i=NR-20;i<=NR;i++){print a[i]}for(i=1;i<35;i++){getline;print}exit}' inp.txt

If its not need to be exact 20/35 lines, you can find 35 lines before and after error, use grep

Code:

grep -C 35 "error" /var/log/messages

Code:

grep -A 35 -B 20 "error" file

is your OS is linux or SunOS ?

if it is linux, you can use -A -B or -C in grep ( gnu grep )