Shell – Diff, show modified line from right file only

Case 1:

awk '!NF {if (++n <= 2) print; next}; {n=0;print}'

Case 2:

awk '!NF {s = s $0 "\n"; n++; next}
     {if (n>1) printf "%s", s; n=0; s=""; print}
     END {if (n>1) printf "%s", s}'

With GNU sed:

sed -n '132 {/^#termcapinfo[[:space:]]*xterm Z0=/q}; $q1'

How it works

• 132 {/^#termcapinfo[[:space:]]*xterm Z0=/q}

  On line 132, check for the regex ^#termcapinfo[[:space:]]*xterm Z0=. If found quit, q, with the default exit code of 0. The rest of the file is skipped.

• $q1

  If we reach the last line, $, then quit with exit code 1: q1.

Efficiency

Since it is not necessary to read past the 132nd line of the file, this version quits as soon as we reach the 132nd line or the end of the file, whichever occurs first:

sed -n '132 {/^#termcapinfo[[:space:]]*xterm Z0=/q; q1}; $q1'

Handling empty files

The version above will return true for empty files. This is because, if the file empty, no commands are executed and the sed exits with the default exit code of 0. To avoid this:

! sed -n '132 {/^#termcapinfo[[:space:]]*xterm Z0=/q1; q}'

Here, the sed command exits with code 0 unless the the desired string is found in which case it exits with code 1 The preceding ! tells the shell to invert this code to get back to the code we want. The ! modifier is supported by all POSIX shells. This version will work even for empty files. (Hat tip: G-Man)