Bash regex finding and replacing

Question

I don't know if this is possible, but can you dynamically alter a find/replace? Basically I have something like this

<3 digit number> <data>

and what I want to do is if data matches the pattern

<word>:<4 digit number>

replace all instances (in the entire file) of <word>: with the line's 3 digit number I.E:

020 Word
021 Word:0001
Replace with 
020 021
021 0210001

Is this doable with AWK or Sed? If not, is it doable in C?

Answer 1

I know this isn't what you asked, but I think the best way to solve this is with a simple Perl script.

#!/usr/bin/perl

$in= "input.txt";
$out= "output.txt";

# Buffer the whole file for replacing:
open(INFILE, $in);
@lines = <INFILE>;
open(INFILE, $in);

# Iterate through each line:
while(<INFILE>) {
  # If the line matches "word:number", replace all instances in the file
  if (/^(\d{3}) (\w+:)\d{4}$/) {
    $num = $1; word = $2;
    s/$word/$num/ foreach @lines;
  }
}

open(OUTFILE, $out);
print OUTFILE foreach @lines;

It looks a lot longer than it really needs to be, because I made it nice and easy-to-read for you.

Answer 2

I hope this time I got you right.

try the stuff below:

#file name:t
kent$  cat t
020 Word
021 Word:0001

#first we find out the replacement, 021 in this case:
kent$  v=$(grep -oP "(\d{3})(?= Word:\d{4})" t|head -n1)

#do replace by sed:
kent$  sed -r "s/Word[:]?/$v/g" t                                                                                                        
020 021 
021 0210001