excluding first and last lines from sed /START/,/END/

Chen Levy picture Chen Levy · Jul 27, 2009 · Viewed 19.3k times · Source

Consider the input:

=sec1=
some-line
some-other-line

foo
bar=baz

=sec2=
c=baz

If I wish to process only =sec1= I can for example comment out the section by:

sed -e '/=sec1=/,/=[a-z]*=/s:^:#:' < input

... well, almost.

This will comment the lines including "=sec1=" and "=sec2=" lines, and the result will be something like:

#=sec1=
#some-line
#some-other-line
#
#foo
#bar=baz
#
#=sec2=
c=baz

My question is: What is the easiest way to exclude the start and end lines from a /START/,/END/ range in sed?

I know that for many cases refinement of the "s:::" claws can give solution in this specific case, but I am after the generic solution here.

In "Sed - An Introduction and Tutorial" Bruce Barnett writes: "I will show you later how to restrict a command up to, but not including the line containing the specified pattern.", but I was not able to find where he actually show this.

In the "USEFUL ONE-LINE SCRIPTS FOR SED" Compiled by Eric Pement, I could find only the inclusive example:

# print section of file between two regular expressions (inclusive)
sed -n '/Iowa/,/Montana/p'             # case sensitive

Answer

Ville Laurikari picture Ville Laurikari · Jul 27, 2009

This should do the trick:

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

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

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

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