Vcsn supports this operator (which it denotes ') eД¬onvert this expression to an automaton: In : a = e.automaton() aДЇinally, convert this automaton back to a simple expression.

Finally, the start- and end-of-input are anchored to make sure the entire input is consumed: ^((?!hede).)*$ДЄs you can see, the input 'ABhedeCD' will fail because on e3, the regex (?!hede) fails (there is 'hede' up ahead!).ДЇWIW, since regular languages (aka rational languages) are closed under complementation, it's always possible to find a regular expression (aka rational expression) that negates another expression. will do that only once, so it is wrapped in a group, and repeated zero or more times: ((?!hede).)*. So, in my example, every empty string is first validated to see if there's no 'hede' up ahead, before a character is consumed by the. Look-arounds are also called zero-width-assertions because they don't consume any characters. (dot) will match any character except a line break. looks ahead to see if there's no substring 'hede' to be seen, and if that is the case (so something else is seen), then the. So a list of n characters will have n+1 empty strings. Before, and after each character, there's an empty string. If the DOT-ALL modifier is not available, you can mimic the same behavior with the character class : /^((?!hede))*$/ДЄ string is just a list of n characters. (where the /./ are the regex delimiters, i.e., not part of the pattern) As mentioned, this is not something regex is 'good' at (or should do), but still, it is possible.ДЄnd if you need to match line break chars as well, use the DOT-ALL modifier (the trailing s in the following pattern): /^((?!hede).)*$/s The regex above will match any string, or line without a line break, not containing the (sub)string 'hede'. You can mimic this behavior by using negative look-arounds: ^((?!hede).)*$ The notion that regex doesn't support inverse matching is not entirely true.

Regular expression not include word