A regex to match a substring that isn't followed by a certain other substring
JavaRegexClojureJava Problem Overview
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.
Java Solutions
Solution 1 - Java
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
Solution 2 - Java
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.
Solution 3 - Java
Use a negative look ahead instead:
\s*(?!\w*(bar)\w*)\w*(foo)\w*\s*
This worked for me, hope it helps. Good luck!
Solution 4 - Java
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
Solution 5 - Java
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.