Matching two words with some characters in between in regular expression

Question

I want to do a match for a string when no abc is followed by some characters (possibly none) and ends with .com.

I tried with the following

Answer 1

Condensing:

Sorry if I did not make myself clear. Just give some examples.
I want def.edu, abc.com, abce.com, eabc.com and
abcAnYTHing.com do not match,
while a.com, b.com, ab.com, ae.com etc. match.

New regex after revised OP examples:
/^(?:(?!abc.*\.com\$|^def\.edu\$).)+\.(?:com|edu)\$/s

use strict;
use warnings;


my @samples = qw/
 
   shouldn't_pass 
   def.edu  abc.com  abce.com eabc.com 
 
   should_pass.com
   a.com    b.com    ab.com   ae.com
   abc.edu  def.com  defa.edu
 /;

my $regex = qr
  /
    ^    # Begin string
      (?:  # Group    

          (?!              # Lookahead ASSERTION
                abc.*\.com$     # At any character position, cannot have these in front of us.
              | ^def\.edu$      # (or 'def.*\.edu$')
           )               # End ASSERTION

           .               # This character passes

      )+   # End group, do 1 or more times

      \.   # End of string check,
      (?:com|edu)   # must be a '.com' or '.edu' (remove if not needed)

    $    # End string
  /sx;


print "\nmatch using   /^(?:(?!abc.*\.com\$|^def\.edu\$).)+\.(?:com|edu)\$/s \n";

for  my $str ( @samples )
{
   if ( $str =~ // ) {
      print "\n"; next;
   }

   if ( $str =~ /$regex/ ) {
       printf ("passed - $str\n");
   }
   else {
       printf ("failed - $str\n");
   }
}

Output:

match using /^(?:(?!abc.*.com$|^def.edu$).)+.(?:com|edu)$/s

failed - shouldn't_pass
failed - def.edu
failed - abc.com
failed - abce.com
failed - eabc.com

passed - should_pass.com
passed - a.com
passed - b.com
passed - ab.com
passed - ae.com
passed - abc.edu
passed - def.com
passed - defa.edu