A regex to match a substring that isn't followed by a certain other substring.

Question

I need a regex that will match blahfooblah but not blahfoobarblah

I want it to match only foo and everything around foo, as long as it isn't followed by bar.

I tried using this: foo.*(?<!bar) which is fairly close, but it matches blahfoobarblah. The negative look behind needs to match anything and not just bar.

The specific language I'm using is Clojure which uses Java regexes under the hood.

EDIT: More specifically, I also need it to pass blahfooblahfoobarblah but not blahfoobarblahblah.

Answer 1

Try:

/(?!.*bar)(?=.*foo)^(\w+)$/

Tests:

blahfooblah            # pass
blahfooblahbarfail     # fail
somethingfoo           # pass
shouldbarfooshouldfail # fail
barfoofail             # fail

Regular expression explanation

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
    .*                       any character except \n (0 or more times
                             (matching the most amount possible))
--------------------------------------------------------------------------------
    bar                      'bar'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    .*                       any character except \n (0 or more times
                             (matching the most amount possible))
--------------------------------------------------------------------------------
    foo                      'foo'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    \w+                      word characters (a-z, A-Z, 0-9, _) (1 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  $                        before an optional \n, and the end of the
                           string

Other regex

If you only want to exclude bar when it is directly after foo, you can use

/(?!.*foobar)(?=.*foo)^(\w+)$/

---

Edit

You made an update to your question to make it specific.

/(?=.*foo(?!bar))^(\w+)$/

New tests

fooshouldbarpass               # pass
butnotfoobarfail               # fail
fooshouldpassevenwithfoobar    # pass
nofuuhere                      # fail

New explanation

(?=.*foo(?!bar)) ensures a foo is found but is not followed directly bar

Answer 2

Use a negative look ahead instead:

\s*(?!\w*(bar)\w*)\w*(foo)\w*\s*

This worked for me, hope it helps. Good luck!

Answer 3

To match a foo following by something that doesn't start with bar, try

foo(?!bar)

Your version with negative lookbehind is effectively "match a foo followed by something that doesn't end in bar". The .* matches all of barblah, and the (?<!bar) looks back at lah and checks that it doesn't match bar, which it doesn't, so the whole pattern matches.

Answer 4

Your specific match request can be matched by:

\w+foo(?!bar)\w+

This will match blahfooblahfoobarblah but not blahfoobarblahblah.

The problem with your regex of foo.*(?<!bar) is the .* after foo. It matches as many of any characters including characters after bar.

Answer 5

You wrote a comment suggesting you like this to work matching all words in a string rather than the whole string itself.

Rather than mashing all of this in a comment, I'm posting it as a new answer.

New Regex

/(?=\w*foo(?!bar))(\w+)/

Sample text

foowithbar fooevenwithfoobar notfoobar foohere notfoobarhere butfooisokherebar notfoobarhere andnofuu needsfoo

Matches

foowithbar fooevenwithfoobar foohere butfooisokherebar needsfoo