Print all lines between two patterns, exclusive, first instance only (in sed, AWK or Perl)

Question

Using sed, AWK (or Perl), how do you print all lines between (the first instance of) two patterns, exclusive of the patterns?¹

That is, given as input:

aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee

Or possibly even:

aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
fff
PATTERN1
ggg
hhh
iii
PATTERN2
jjj

I would expect, in both cases:

bbb
ccc
ddd

---

¹ _{A number of users voted to close this question as a duplicate of this one. In the end, I provided a gist that proves they are different. The question is also superficially similar to a number of others, but there is no exact match, and none of them are of high quality, and, as I believe that this specific problem is the one most commonly faced, it deserves a clear formulation, and a set of correct, clear answers.}

Answer 1

If you have GNU sed (tested using version 4.7 on Mac OS X), the simplest solution could be:

sed '0,/PATTERN1/d;/PATTERN2/Q'

Explanation:

• The d command deletes from line 1 to the line matching /PATTERN1/ inclusive.
• The Q command then exits without printing on the first line matching /PATTERN2/.

If the file has only once instance of the pattern, or if you don't mind extracting all of them, and you want a solution that doesn't depend on a GNU extension, this works:

sed -n '/PATTERN1/,/PATTERN2/{//!p}'

Explanation:

• Note that the empty regular expression // repeats the last regular expression match.

Answer 2

With awk (assumes that PATTERN1 and PATTERN2 are always present in pairs and either of them do not occur inside a pair)

$ cat ip.txt
aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
fff
PATTERN1
ggg
hhh
iii
PATTERN2
jjj

$ awk '/PATTERN2/{exit} f; /PATTERN1/{f=1}' ip.txt
bbb
ccc
ddd

• /PATTERN1/{f=1} set flag if /PATTERN1/ is matched
• /PATTERN2/{exit} exit if /PATTERN2/ is matched
• f; print input line if flag is set

Generic solution, where the block required can be specified

$ awk -v b=1 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
bbb
ccc
ddd
$ awk -v b=2 '/PATTERN2/ && c==b{exit} c==b; /PATTERN1/{c++}' ip.txt
2
46

Answer 3

This might work for you (GNU sed);

sed -n '/PATTERN1/{:a;n;/PATTERN2/q;p;$!ba}' file

This prints only the lines between the first set of delimiters, or if the second delimiter does not exist, to the end of the file.

Answer 4

I attempted twice to answer, but the questions switched hold/duplicate statuses..

Borrowing input from @Sundeep and adding the answer which I shared in the question comments.

Using awk

awk -v x=0 -v y=1 ' /PATTERN1/&&y { x=1;next } /PATTERN2/&&y { x=0;y=0; next } x ' file

with Perl

perl -0777 -ne ' while( /PATTERN1.*?\n(.+?)^[^\n]*?PATTERN2/msg ) { print $1 if $x++ <1 } '

Results:

$ cat ip.txt
aaa
PATTERN1
bbb
ccc
ddd
PATTERN2
eee
PATTERN1
2
46
PATTERN2
xyz

$

$ awk -v x=0 -v y=1 ' /PATTERN1/&&y { x=1;next } /PATTERN2/&&y { x=0;y=0; next } x ' ip.txt
bbb
ccc
ddd

$ perl -0777 -ne ' while( /PATTERN1.*?\n(.+?)^[^\n]*?PATTERN2/msg ) { print $1 if $x++ <1 } ' ip.txt
bbb
ccc
ddd

$

To make it generic

awk here y is the input

awk -v x=0 -v y=2 ' /PATTERN1/ { x++;next } /PATTERN2/ { if(x==y) exit } x==y ' ip.txt
2
46

perl check ++$x against the occurence.. here it is 2

perl -0777 -ne ' while( /PATTERN1.*?\n(.+?)^[^\n]*?PATTERN2/msg ) { print $1 if ++$x==2 } ' ip.txt
2
46

Answer 5

Adding more solutions(possible ways here, for fun :) and not at all claiming that these are better than usual ones) All tested and written in GNU awk. Also tested with given examples only.

1st Solution:

awk -v RS="" -v FS="PATTERN2" -v ORS="" '$1 ~ /\nPATTERN1\n/{sub(/.*PATTERN1\n/,"",$1);print $1}' Input_file

2nd solution:

awk -v RS="" -v ORS="" 'match($0,/PATTERN1[^(PATTERN2)]*/){val=substr($0,RSTART,RLENGTH);gsub(/^PATTERN1\n|^$\n/,"",val);print val}' Input_file

3rd solution:

awk -v RS="" -v OFS="\n" -v ORS="" 'sub(/PATTERN2.*/,"") && sub(/.*PATTERN1/,"PATTERN1"){$1=$1;sub(/^PATTERN1\n/,"")} 1' Input_file

In all above codes output will be as follows.

bbb
ccc
ddd

Answer 6

Using GNU sed:

sed -nE '/PATTERN1/{:s n;/PATTERN2/q;p;bs}'

-n will prune all but lines between PATTERN1 and PATTERN2 including both, because there will be p printout command. every sed range check if it's true will execute only one the next, so {} grouping is mandated.. Drop PATTERN1 by n command (means next), if reach the first PATTERN2 outrightly quit otherwise print the line then and continue the next line within that boundary.