mjhjmtu

How do I write a sed script that scans the input file for "start" and finds the line containing "next" and displays the following line? Something like this:

[user]$ cat test.txt
start
next
This line should print
Ignore this


[user]$ display.sed test.txt
This line should print

[user]$ cat test1.txt
Ignore this
next
Ignore this
start
Ignore this
next
This line should print
Ignore this
next
Too late so ignore this too
start
Ignore this too

[user]$ display.sed test1.txt
This line should print

You could use a range (from 1st occurrence of start to end of file) and delete all lines in that range that don't match next. If a line matches, read in the next line, print it then quit:

sed -n '/start/,$
/next/!d;n;p;q
' infile

I guess what you actually want is a file display.sed with the following content:

#!/bin/sed -nf

/start/,$
/next/!d;n;p;q

Using awk would be more appropriate for such case cause it should work even on non-GNU awk implementations:

awk '/^start/ f=1 f && /^next/ && getline nl>0 print nl; exit ' test.txt

• /^start/ f=1 - set active flag f=1 on encountering line start




• f && /^next/ && getline nl>0 - on encountering next line (with previously matched start line - ensured by active f flag) - check if the next needed line exists with getline nl>0




• nl ("needed line") - contains the line following line next

The output (for your current input content):

This line should print

What about grep? I will add explanation, if this command does, what you want.

grep -Pzo '(?s)start.*?nextnK.*?n' input.txt

Input (your both examples merged)

start
next
This line should print
Ignore this
Ignore this
next
Ignore this
start
Ignore this
next
This line should print
Ignore this
next
Too late so ignore this too 
start
Ignore this too 

Output

This line should print
This line should print