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Display few rows before and after the pattern is found

 
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# 1  
Old 05-13-2009
Question Display few rows before and after the pattern is found
hi,
i'm trying to write a script in ksh that checks the build log for errors. I use egrep to serch for a pattern list. When the pattern is found i want to print 2 rows before and 2 rows after the line where the patern is found, but i don't know how to do it.
# 2  
Old 05-13-2009
Below shell script will do the needful

Code:
>output
grep -n <pattern> <filename> | cut -d':' -f1 > line_nos
for i in `cat line_nos `
do
head -$(($i-1)) <filename>|tail -1 >> output
head -$(($i-2)) <filename>|tail -1 >> output
head -$(($i+1)) <filename>|tail -1 >> output
head -$(($i+2)) <filename>|tail -1 >> output
done

s kulshrestha
Last edited by Yogesh Sawant; 06-01-2010 at 10:50 AM.. Reason: added code tags
# 3  
Old 05-13-2009
Quote:
Originally Posted by lavinia_f
hi,
i'm trying to write a script in ksh that checks the build log for errors. I use egrep to serch for a pattern list. When the pattern is found i want to print 2 rows before and 2 rows after the line where the patern is found, but i don't know how to do it.
Use grep if your grep version supports the -A and -B option.

Regards
# 4  
Old 05-13-2009
It worked. Thank you Smilie
 
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