How to negate the whole regex?

 

I have a regex, for example (ma|(t){1}) . It matches ma and t and doesn't match bla .

I want to negate the regex, thus it must match bla and not ma and t , by adding something to this regex . I know I can write bla , the actual regex is however more complex.

Use negative lookaround: (?! pattern )

Positive lookarounds can be used to assert that a pattern matches. Negative lookarounds is the opposite: it's used to assert that a pattern DOES NOT match. Some flavor supports assertions; some puts limitations on lookbehind, etc.

Links to regular-expressions.info

  • Lookahead and Lookbehind Zero-Width Assertions
  • Flavor comparison

    • See also

  • How do I convert CamelCase into human-readable names in Java?
  • Regex for all strings not containing a string?
  • A regex to match a substring that isn't followed by a certain other substring.

    • More examples

    These are attempts to come up with regex solutions to toy problems as exercises; they should be educational if you're trying to learn the various ways you can use lookarounds (nesting them, using them to capture, etc):

  • codingBat plusOut using regex
  • codingBat repeatEnd using regex
  • codingbat wordEnds using regex

    • Assuming you only want to disallow strings that match the regex completely (ie, mmbla is okay, but mm isn't), this is what you want: 

    ^(?!(?:m{2}|t)$).*$

    (?!(?:m{2}|t)$) is a negative lookahead; it says "starting from the current position, the next few characters are not mm or t , followed by the end of the string." The start anchor ( ^ ) at the beginning ensures that the lookahead is applied at the beginning of the string. If that succeeds, the .* goes ahead and consumes the string.

    FYI, if you're using Java's matches() method, you don't really need the the ^ and the final $ , but they don't do any harm. The $ inside the lookahead is required, though. 

    b(?=w)(?!(ma|(t){1}))b(w*)

    this is for the given regex.
    the b is to find word boundary.
    the positive look ahead (?=w) is here to avoid spaces.
    the negative look ahead over the original regex is to prevent matches of it.
    and finally the (w*) is to catch all the words that are left.
    the group that will hold the words is group 3.
    the simple (?!pattern) will not work as any sub-string will match
    the simple ^(?!(?:m{2}|t)$).*$ will not work as it's granularity is full lines

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