Extract K-th Line from Chunks Using Sed/AWK/Perl - linux

Question

I have some data that looks like this. It comes in chunk of four lines. Each chunk starts with a # character.
#SRR037212.1 FC30L5TAA_102708:7:1:741:1355 length=27
AAAAAAAAAAAAAAAAAAAAAAAAAAA
+SRR037212.1 FC30L5TAA_102708:7:1:741:1355 length=27
::::::::::::::::::::::::;;8
#SRR037212.2 FC30L5TAA_102708:7:1:1045:1765 length=27
TATAACCAGAAAGTTACAAGTAAACAC
+SRR037212.2 FC30L5TAA_102708:7:1:1045:1765 length=27
88888888888888888888888888
What I want to do is to extract last line of each chunk. Yielding:
::::::::::::::::::::::::;;8
888888888888888888888888888
Note that the last line of the chunk may contain any standard ASCII character
including #.
Is there an effective one-liner to do it?

Answer 1

The following sed command will print the 3rd line after the pattern:
sed -n '/^#/{n;n;n;p}' file.txt

Answer 2

If there are no blank lines:
perl -ne 'print if $. % 4 == 0' file

Answer 3

$ awk 'BEGIN{RS="#";FS="\n"}{print $4 } ' file
::::::::::::::::::::::::;;8
88888888888888888888888888
If you always have those 4 lines in a chunk, some other ways
$ ruby -ne 'print if $.%4==0' file
::::::::::::::::::::::::;;8
88888888888888888888888888
$ awk 'NR%4==0' file
::::::::::::::::::::::::;;8
88888888888888888888888888
It also seems like your line is always after the line that start with "+", so
$ awk '/^\+/{getline;print}' file
::::::::::::::::::::::::;;8
88888888888888888888888888
$ ruby -ne 'gets && print if /^\+/' file
::::::::::::::::::::::::;;8
88888888888888888888888888

Answer 4

This prints the lines before lines that starts with #, and also the last line. It can work with non uniform sized chunks, but assumes that only a chunk leading line starts with #.
sed -ne '1d;$p;/^#/!{x;d};/^#/{x;p}' file
Some explanation is in order:
First you don't need the first line so delete it 1d
Next you always need the last line, so print it $p
If you don't have a match swap it into the hold buffer and delete it x;d
If you do have match swap it out of the hold buffer, and print it x;p

Answer 5

This works similarly to dogbane's answer
awk '/^#/ {mark = NR} NR == mark + 3 {print}' inputfile
And, like that answer, will work regardless of the number of lines in each chunk (as long as there are at least 4).
The direct analog to that answer, however, would be:
awk '/^#/ {next; next; next; print}' inputfile

Answer 6

this can be done using grep easily
grep -A 1 '^#' ./infile

Answer 7

This might work for you (GNU sed):
sed '/^#/,+2d' file