Here's what I did: 
#!/usr/bin/perl -w

use strict;

my @lines = (
    'Bob Smith bsmith 00001234567 01/01/1986 00:00:00',
    'Mary Ann Doe mdoe 00001234568 01/01/1986 00:00:00 00001234563 01/
+01/1986 00:00:00',
    'Gilligan Q Smith gsmith 00001234569 01/01/1986 00:00:00');

foreach my $line (@lines) {
    $line =~ s/(\s[a-z]+\s0{4})/,$1/g;
    print "New line:\n'$line'\n";
}

[download]
I didn't use \w because that goes for alphanumerics. I matched the whole login name with the character class [a-z]+, with one whitespace character before and after it. I didn't need to match the rest of the string, because that particular combination only occurs once in your dataset.

If I were doing this on my own, I might look into split. I like split.

Split on the token (you know it's an n-digit number): 

my ($user, $token, $login) = split(/\s(\d{11})\s/, $line, 2);
(my $username = $user) =~ s/\s(\w+)$//;        # untested construct
print "$username, claiming to be $user logged on with token $token, at
+ $login.\n";

[download]
Then again, the regex seems pretty straightforward.

In reply to Re: Backtracking for substitutions by chromatic
in thread Backtracking for substitutions by Anonymous Monk

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