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+)/ an

Answer 1

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 2

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 3

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

Answer 4

GNU awk

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