grep -n lines before and after

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# 1  
Old 01-21-2008
grep -n lines before and after


is it possible to grep a pattern that will include the "n" lines before and after the line where the pattern have been found?


#this contains the test.file

then a grep command to search the word "line3"
and the output should be 1 (or n) line before that line and 1 (or n) line "after" that line.

dessired output of the grep command


Thanks in advance.
# 2  
Old 01-21-2008
If you have GNU grep, then it is possible. From man grep

       -A NUM, --after-context=NUM
              Print NUM lines  of  trailing  context  after  matching  lines.
              Places  a  line  containing  --  between  contiguous  groups of

       -B NUM, --before-context=NUM
              Print  NUM  lines  of  leading  context  before matching lines.
              Places a  line  containing  --  between  contiguous  groups  of

# 3  
Old 01-21-2008
# 4  
Old 07-03-2009

There is a very simple way of doing this. Lets say you have a file test.txt having 'view' in some line. Say, you wish to get 5 lines above and below the line containing 'view' and output to a file say test_one.txt use the following:

grep -C 5 "view" test.txt > test_one.txt

# 5  
Old 07-03-2009
Please note: a lot of these examples ONLY work with GNU tools, not all versions of grep.
# 6  
Old 07-05-2009
hope below perl can help you some

sub lines_grep{
		print @tmp;	
		if($#tmp < $line-1){
			push @tmp, $_;
			shift @tmp;
			push @tmp, $_;
	if ($flag==1){
		print $_ ;

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