Bash: Split text-file into words with non-alphanumeric characters as delimiters

Question

Lets say "textfile" contains the following:

lorem$ipsum-is9simply the.dummy text%of-printing

and that you want to print each word on a separate line. However, words should be defined not only by spaces, but by all non-alphanumeric characters. So the results should look like:

 lorem
 ipsum  
 is9simply  
 the  
 dummy  
 text  
 of  
 printing

How can I accomplish this using the Bash shell?

---

Some notes:

• This is not a homework question.

• The simpler case when the words should be determined only by spaces, is easy. Just writing:

  for i in `cat textfile`; do echo $i; done;
  

  will do the trick, and return:

   lorem$ipsum-is9simply
   the.dummy
   text%of-printing
  

  For splitting words by non-alphanumeric characters I have seen solutions that use the IFS environmental variable (links below), but I would like to avoid using IFS for two reasons: 1) it would require (I think) setting the IFS to a long list of non-alphanumeric characters. 2) I find it kind of ugly.

• Here are the two related Q&As I found
  http://stackoverflow.com/questions/918886/split-string-based-on-delimiter-in-bash
  http://stackoverflow.com/questions/1975849/split-line-into-words-in-bash

Answer 1

Use the tr command:

tr -cs 'a-zA-Z0-9' '\n' <textfile

The '-c' is for the complement of the specified characters; the '-s' squeezes out duplicates of the replacements; the 'a-zA-Z0-9' is the set of alphanumerics; the '\n' is the replacement character (newline). You could also use a character class which is locale sensitive (and may include more characterss than the list above):

tr -cs '[:alnum:]' '\n' <textfile

Answer 2

$ awk -f splitter.awk < textfile

$ cat splitter.awk
{
  count0 = split($0, asplit, "[^a-zA-Z0-9]")
  for(i = 1; i <= count0; ++i) { print asplit[i] }
}