[SOLVED] Return Values Below String Pattern Match in Number Range

Issue

I need to match and return the values below the number range 12-00 in the first line/row (UTC) of the following text file:

UTC  06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 00 01 02 03 04 05 06 
TMP  54 53 52 50 49 48 47 47 47 48 48 48 48 48 47 45 44 43 43 41 40 39 38 37 36

So that is, matching 12 13 14 15 16 17 18 19 20 21 22 23 00 in Line 1 and returning 47 47 47 48 48 48 48 48 47 45 44 43 43 from Line 2.

My Attempt:

cat some.text.file | head -n 3 | grep -A 1 '12.*.00' | tail -n 1

Result:

TMP  54 53 52 50 49 48 47 47 47 48 48 48 48 48 47 45 44 43 43 41 40 39 38 37 36

Expected Result:

12 13 14 15 16 17 18 19 20 21 22 23 00
47 47 47 48 48 48 48 48 47 45 44 43 43

Solution

This can be done in a single awk:

awk 'NR == 1 {for (i=1; i<=NF; ++i) if ($i == "12") start = i; else if ($i == "00") stop = i} {for (i=start; i<=stop; ++i) printf "%s", $i (i < stop ? OFS : ORS)}' file

12 13 14 15 16 17 18 19 20 21 22 23 00
47 47 47 48 48 48 48 48 47 45 44 43 43

A more readable version:

awk 'NR == 1 {
   for (i=1; i<=NF; ++i)
      if ($i == "12")
         start = i
      else if ($i == "00")
         stop = i
}
{
   for (i=start; i<=stop; ++i)
      printf "%s", $i (i < stop ? OFS : ORS)
}' file

Answered By - anubhava

Answer Checked By - Mildred Charles (WPSolving Admin)