Exclusive Or in Regular Expression

Question

Looking for a bit of regex help. I'd like to design an expression that matches a string with "foo" OR "bar", but not both "foo" AND "bar"

If I do something like...

/((foo)|(bar))/

It'll match "foobar". Not what I'm looking for. So, how can I make regex match only when one term or the other is present?

Thanks!

Escaut

Would foofoobar be a match because it contains "foo" and "foobar"? How about "foonbar"? Could you provide examples of matches and non-matches? – Calash

Matches: "foo", "bar" nonmatches: "foofoo" "barfoo" "foobarfoo" "barbar" "barfoofoo" – Escaut

If you don't want "foofoo" to match, then you're not really talking about an exclusive or. – Seat

Answer 1

You can do this with a single regex but I suggest for the sake of readability you do something like...

(/foo/ and not /bar/) || (/bar/ and not /foo/)

Imperfection

Indeed, I'm pretty sure I would put the XOR logic into the code itself, and not in the regexp. – Sorbian

Or even better, /foo/ xor /bar/, if your language has an XOR operator. (Perl does.) – Seat

@Ralf This is not a single expression, it is two expressions joined with a logical OR operator. – Imperfection

Answer 2

This is what I use:

/^(foo|bar){1}$/

See: http://www.regular-expressions.info/quickstart.html under repetition

Ohmmeter

Much more elegant solution than the accepted answer, especially when you got more than 2 cases.. – Thai

Why did you add {1}, what does that mean? – Liege

This is wrong, it only means foo or bar should be matched only once. – Mesocratic

I agree with @Mesocratic this is not an XOR. It only checks if the whole string is "foo" or "bar" – Tippett

You don’t need the {1}, because that indicates it should be repeated 1 times. It’s sufficient to have /^(foo|bar)$/. The reason this regex works is that you’ve used ^ and $; it has nothing to do with repetition. – Southworth

Answer 3

If your regex language supports it, use negative lookaround:

(?<!foo|bar)(foo|bar)(?!foo|bar)

This will match "foo" or "bar" that is not immediately preceded or followed by "foo" or "bar", which I think is what you wanted.

It's not clear from your question or examples if the string you're trying to match can contain other tokens: "foocuzbar". If so, this pattern won't work.

Here are the results of your test cases ("true" means the pattern was found in the input):

foo: true
bar: true
foofoo: false
barfoo: false
foobarfoo: false
barbar: false
barfoofoo: false

Adda

Answer 4

This will take 'foo' and 'bar' but not 'foobar' and not 'blafoo' and not 'blabar':

/^(foo|bar)$/

^ = mark start of string (or line)
$ = mark end of string (or line)

This will take 'foo' and 'bar' and 'foo bar' and 'bar-foo' but not 'foobar' and not 'blafoo' and not 'blabar':

/\b(foo|bar)\b/

\b = mark word boundry

Liege

Answer 5

You haven't specified behaviour regarding content other than "foo" and "bar" or repetitions of one in the absence of the other. e.g., Should "food" or "barbarian" match?

Assuming that you want to match strings which contain only one instance of either "foo" or "bar", but not both and not multiple instances of the same one, without regard for anything else in the string (i.e., "food" matches and "barbarian" does not match), then you could use a regex which returns the number of matches found and only consider it successful if exactly one match is found. e.g., in Perl:

@matches = ($value =~ /(foo|bar)/g)  # @matches now hold all foos or bars present
if (scalar @matches == 1) {          # exactly one match found
  ...
}

If multiple repetitions of that same target are allowed (i.e., "barbarian" matches), then this same general approach could be used by then walking the list of matches to see whether the matches are all repeats of the same text or if the other option is also present.

Inaugurate

Answer 6

You might want to consider the ? conditional test.

(?(?=regex)then|else)

Regular Expression Conditionals

Arlo

Answer 7

If you want a true exclusive or, I'd just do that in code instead of in the regex. In Perl:

/foo/ xor /bar/

But your comment:

Matches: "foo", "bar" nonmatches: "foofoo" "barfoo" "foobarfoo" "barbar" "barfoofoo"

indicates that you're not really looking for exclusive or. You actually mean "Does /foo|bar/ match exactly once?"

my $matches = 0;
while (/foo|bar/g) {
  last if ++$matches > 1;
}

my $ok = ($matches == 1)

Seat

Answer 8

I know this is a late entry, but just to help others who may be looking:

(/b(?:(?:(?!foo)bar)|(?:(?!bar)foo))/b)

Wast

Answer 9

I'd use something like this. It just checks for space around the words, but you could use the \b or \B to check for a border if you use \w. This would match " foo " or " bar ", so obviously you'd have to replace the whitespace as well, just in case. (Assuming you're replacing anything.)

/\s((foo)|(bar))\s/

Hipbone

Answer 10

I don't think this can be done with a single regular expression. And boundaries may or may not work depending on what you're matching against.

I would match against each regex separately, and do an XOR on the results.

foo = re.search("foo", str) != None
bar = re.search("bar", str) != None
if foo ^ bar:
    # do someting...

Tyner

Answer 11

I tried with Regex Coach against:

x foo y
x bar y
x foobar y

If I check the g option, indeed it matches all three words, because it searches again after each match.
If you don't want this behavior, you can anchor the expression, for example matching only on word boundaries:

\b(foo|bar)\b

Giving more context on the problem (what the data looks like) might give better answers.

Tager

Answer 12

\b(foo)\b|\b(bar)\b

And use only the first capture group.

Mellow

Answer 13

Using the word boundaries, you can get the single word...

me@home ~  
$ echo "Where is my bar of soap?" | egrep "\bfoo\b|\bbar\b"  
Where is my bar of soap?  

me@home ~  
$ echo "What the foo happened here?" | egrep "\bfoo\b|\bbar\b"  
What the foo happened here?  

me@home ~  
$ echo "Boy, that sure is foobar\!" | egrep "\bfoo\b|\bbar\b"  

Faulk

Answer 14

for starters, it helps to have a good test case:

apple
banana
orange
apple and banana
apple and orange
banana and apple
banana and orange
orange and apple
orange and banana
apple and banana and orange
orange and banana and orange
none of the above
apple and microsoft

https://regex101.com/r/ebpyjX/1

first, ^(?P<XOR>(?P<apple_not_banana>(?=.*apple)(?!.*banana))|(?P<banana_not_apple>(?!.*apple)(?=.*banana))).*$ is reads as"apple NOT banana, OR, not apple, banana" works directly as XOR but would potentially become long with many subterms:

one and a halfth, ^(?P<XOR>(?=.*apple)(?!.*banana)(?!.*orange)|(?!.*apple)(?=.*banana)(?!.*orange)|(?!.*apple)(?!.*banana)(?=.*orange)).*$ reads as "apple not banana not orange; not apple, banana, not orange; not apple, not banana, orange" and is an example of the three-valued xor, via enumeration of the product:

one and three quarters ^(?P<XOR>(?!(?P<exclusive>(.*apple|.*banana|.*orange){2,}))(?=.*(?P<fruit>apple|banana|orange)).*$) reads as, "NOT (apple, banana, or orange) 2 or more times, and apple or banana or orange" could scale better in terms of length to larger numbers of terms because it groups logic more effectively, EDIT: however, in practice searching NLM UMLS, I discovered this approach may cause catastrophic backtracking.

second, (?P<xor_is_or_and_nand>^(?P<nand_apple_banana>(?!(?:(?=.*apple)(?=.*banana))))(?P<apple_or_banana>(?=.*apple|.*banana)).*$) reads as "NOT apple AND banana, AND apple OR banana" shows how XOR is OR + NAND i.e. in lisp this would look like (and (or apple banana) (nand apple banana)):

finally, (?P<not_with_xor>^(?P<not_microsoft>(?!.*microsoft))(?P<nand_apple_banana>(?!(?:(?=.*apple)(?=.*banana))))(?P<apple_or_banana>(?=.*apple|.*banana)).*$) reads "NOT microsoft, AND NOT apple AND banana, AND apple OR banana" shows how to ensure coexistence of other negations with the xor query:

Dhyana

Some explanation of these reges would be useful, rather than just showing them with screenshots of their testing. What are all the named capture groups for? Given that the question and it accepted answer is over 15 years old, any new answer should focus on giving new information and giving new insights into the topic. – Surfacetosurface

@Surfacetosurface if you think the accepted answer meets your high standards, then you're welcome to use that! – Dhyana

anyway, named groups are to explain how they work, and for maintainability, readability – Dhyana

This site aims to provide high quality answers for future readers. Many of us seek to maintain the usefulness of the questions and answers here. For the reasons I gave above, I think this answer could be much improved. – Surfacetosurface

there you go, i wrote them out, enjoy your c# – Dhyana

Adrian isn't being negative, he's providing constructive feedback. There is an expectation on this site that the community will always focus on high standards, so they are not just Adrian's high standard. – Mummify

The OP has asked about XOR only. You've provided information that goes beyond what's being asked. It's reducing the usefulness of he answer. – Mummify