[ACCEPTED]-excluding first and last lines from sed /START/,/END/-text-manipulation

Accepted answer
Score: 41

This should do the trick:

sed -e '/=sec1=/,/=sec2=/ { /=sec1=/b; /=sec2=/b; s/^/#/ }' < input

This matches between 10 sec1 and sec2 inclusively and then just 9 skips the first and last line with the b command. This 8 leaves the desired lines between sec1 and 7 sec2 (exclusive), and the s command adds 6 the comment sign.

Unfortunately, you do need 5 to repeat the regexps for matching the delimiters. As 4 far as I know there's no better way to do 3 this. At least you can keep the regexps 2 clean, even though they're used twice.

This 1 is adapted from the SED FAQ: How do I address all the lines between RE1 and RE2, excluding the lines themselves?

Score: 14

If you're not interested in lines outside 8 of the range, but just want the non-inclusive 7 variant of the Iowa/Montana example from 6 the question (which is what brought me here), you 5 can write the "except for the first and 4 last matching lines" clause easily enough 3 with a second sed:

sed -n '/PATTERN1/,/PATTERN2/p' < input | sed '1d;$d'

Personally, I find this 2 slightly clearer (albeit slower on large 1 files) than the equivalent

sed -n '1,/PATTERN1/d;/PATTERN2/q;p' < input

Score: 7

Another way would be

sed '/begin/,/end/ {
       /end/ !p

/begin/n -> skip over the 3 line that has the "begin" pattern
/end/ !p -> print 2 all lines that don't have the "end" pattern

Taken 1 from Bruce Barnett's sed tutorial http://www.grymoire.com/Unix/Sed.html#toc-uh-35a

Score: 2

I've used:

sed '/begin/,/end/{/begin\|end/!p}'

This will search all the lines 2 between the patterns, then print everything 1 not containing the patterns

Score: 1

you could also use awk

awk '/sec1/{f=1;print;next}f && !/sec2/{ $0="#"$0}/sec2/{f=0}1' file


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