awk extract multiple groups from each line

Question

How do I perform action on all matching groups when the pattern matches multiple times in a line?

To illustrate, I want to search for /Hello! (\d+)/ and use the numbers, for example, print them out or sum them, so for input

abcHello! 200 300 Hello! Hello! 400z3
ads
Hello! 0

If I decided to print them out, I'd expect the output of

200
400
0

Answer 1

GNU awk

awk 'BEGIN{ RS="Hello! ";}
{
    gsub(/[^0-9].*/,"",$1)
    if ($1 != ""){ 
        print $1 
    }
}' file

Answer 2

This is gawk syntax. It also works for patterns when there's no fixed text that can work as a record separator and doesn't match over linefeeds:

 {
     pattern = "([a-g]+|[h-z]+)"
     while (match($0, pattern, arr))
     {
         val = arr[1]
         print val
         sub(pattern, "")
     }
 }

Answer 3

This is a simple syntax, and every awk (nawk, mawk, gawk, etc) can use this.

{
    while (match($0, /Hello! [0-9]+/)) {
        pattern = substr($0, RSTART, RLENGTH);
        sub(/Hello! /, "", pattern);
        print pattern;
        $0 = substr($0, RSTART + RLENGTH);
    }
}

Answer 4

There is no gawk function to match the same pattern multiple times in a line. Unless you know exactly how many times the pattern repeats.

Having this, you have to iterate "manually" on all matches in the same line. For your example input, it would be:

{
  from = 0
  pos = match( $0, /Hello! ([0-9]+)/, val )
  while( 0 < pos )
  {
    print val[1]
    from += pos + val[0, "length"]
    pos = match( substr( $0, from ), /Hello! ([0-9]+)/, val )
  }
}

If the pattern shall match over a linefeed, you have to modify the input record separator - RS