How can I print 2 lines if the second line contains the same match as the first line?

Question

Let's say I have a file with several million lines, organized like this:

@1:N:0:ABC
XYZ

@1:N:0:ABC
ABC

I am trying to write a one-line grep/sed/awk matching function that returns both lines if the NCCGGAGA line from the first line is found in the second line.

When I try to use grep -A1 -P and pipe the matches with a match like '(?<=:)[A-Z]{3}', I get stuck. I think my creativity is failing me here.

Answer 1

With awk

$ awk -F: 'NF==1 && $0 ~ s{print p ORS $0} {s=$NF; p=$0}' ip.txt
@1:N:0:ABC
ABC

• -F: use : as delimiter, makes it easy to get last column
• s=$NF; p=$0 save last column value and entire line for printing later
• NF==1 if line doesn't contain :
• $0 ~ s if line contains the last column data saved previously
  • if search data can contain regex meta characters, use index($0,s) instead to search literally
• note that this code assumes input file having line containing : followed by line which doesn't have :

With GNU sed (might work with other versions too, syntax might differ though)

$ sed -nE '/:/{N; /.*:(.*)\n.*\1/p}' ip.txt
@1:N:0:ABC
ABC

• /:/ if line contains :
• N add next line to pattern space
• /.*:(.*)\n.*\1/ capture string after last : and check if it is present in next line

again, this assumes input like shown in question.. this won't work for cases like

@1:N:0:ABC
@1:N:0:XYZ
XYZ

Answer 2

This might work for you (GNU sed):

sed -n 'N;/.*:\(.*\)\n.*\1/p;D' file

Use grep-like option -n to explicitly print lines. Read two lines into the pattern space and print both if they meet the requirements. Always delete the first and repeat.