bash grep newline

Question

[Editorial insertion: Possible duplicate of the same poster\'s earlier question?]

Hi, I need to extract from the file:

first
second
         


        

Answer 1

grep -A1 "second" | grep -B1 "third" works nicely, and if you have multiple matches it will even get rid of the original -- match delimiter

Answer 2

you could use

$ grep -1 third filename

this will print a string with match and one string before and after. Since "third" is in the last string you get last two strings.

Answer 3

So you just don't want the line containing "first"? -v inverts the grep results.

$ echo -e "first\nsecond\nthird\n" | grep -v first
second
third

Answer 4

grep -v '^first' filename

Where the -v flag inverts the match.

Answer 5

Instead of grep, you can use pcregrep which supports multiline patterns

pcregrep -M 'second\nthird' file

-M allows the pattern to match more than one line.

Answer 6

I don't really understand what do you want to match. I would not use grep, but one of the following:

tail -2 file         # to get last two lines
head -n +2 file      # to get all but first line
sed -e '2,3p;d' file # to get lines from second to third

(not sure how standard it is, it works in GNU tools for sure)