Regex: qualifying by not matching a pattern

Question

I have this pattern:
(.[^(?<=d)(?=d)t].*error.*|.{0,2}error.*)and these Strings to match:
1 "dtd validation error"
2 "dtd error"
3 "error"
4 "foobar error"
5 "dt derror"The pattern fails to match #1 and #2, which is good. It succeeds in matching #3 and #4, which is also good. However, it fails to match #5, which is not what I expected.
According to me, the second half of the pattern means find 0, 1 or 2 chars of any type, followed by the word 'error', followed by 0 or more chars of any type. I don't expect #5 to be matched by this one, so that's okay.
However, the first half of the pattern seems to say find one char of any type, followed by a char that is not 't' if prefixed and suffixed with 'd', followed by 0 or more chars of any type, followed by the word 'error', followed by 0 or more chars of any type.
The 't' in #5 is prefixed with a 'd' alright, but it obviously isn't suffixed with a 'd' (as I see it, it's suffixed with a whitespace). How come that the first half of the pattern seems to recognize that it is? Or is there another problem with my expression?

Answer 1

Why not create the opposite pattern? It's easier..
     String pattern = ".*dtd.*error.*";
     String t1 = "dtd validation error";
     String t2 = "dtd error";
     String t3 = "error";
     String t4 = "foobar error";
     String t5 = "dt derror";
     System.out.println(!Pattern.matches(pattern, t1));
     System.out.println(!Pattern.matches(pattern, t2));
     System.out.println(!Pattern.matches(pattern, t3));
     System.out.println(!Pattern.matches(pattern, t4));
     System.out.println(!Pattern.matches(pattern, t5));/Kaj