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!

Would foofoobar be a match because it contains "foo" and "foobar"? How about "foonbar"? Could you provide examples of matches and non-matches? — Thomas Owens, Oct 29 '08 at 15:11
Matches: "foo", "bar" nonmatches: "foofoo" "barfoo" "foobarfoo" "barbar" "barfoofoo" — SocialCensus, Oct 29 '08 at 15:29
If you don't want "foofoo" to match, then you're not really talking about an exclusive or. — cjm, Oct 29 '08 at 17:52

score 45 · Answer 1 · answered Dec 02 '11 at 19:57

45

This is what I use:

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

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

answered Dec 02 '11 at 19:57

Tristan

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Much more elegant solution than the accepted answer, especially when you got more than 2 cases.. – Manuel Arwed Schmidt Sep 28 '15 at 14:48
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Why did you add `{1}`, what does that mean? – oriadam Jun 05 '16 at 10:35
9

This is wrong, it only means `foo` or `bar` should be matched only once. – Karl Jul 05 '16 at 01:56
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I agree with @Karl this is not an XOR. It only checks if the whole string is "foo" or "bar" – ecdani Aug 29 '19 at 07:59
1

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. – chharvey Jan 06 '22 at 01:38

score 20 · Answer 2 · answered Oct 29 '08 at 16:32

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

score 10 · Accepted Answer · answered Oct 29 '08 at 15:11

10

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/)

answered Oct 29 '08 at 15:11

Ed Guiness

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Indeed, I'm pretty sure I would put the XOR logic into the code itself, and not in the regexp. – Pistos Oct 29 '08 at 16:18
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Or even better, /foo/ xor /bar/, if your language has an XOR operator. (Perl does.) – cjm Oct 29 '08 at 17:47
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@Ralf This is not a single expression, it is two expressions joined with a logical OR operator. – Ed Guiness Feb 23 '18 at 11:02

oriadam · Answer 4 · 2018-01-29T15:19:37.200

8

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

edited Jan 29 '18 at 15:19

answered Sep 09 '14 at 10:46

oriadam

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score 3 · Answer 5 · answered Oct 29 '08 at 16:59

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.

score 2 · Answer 6 · answered Oct 29 '08 at 15:56

2

You might want to consider the ? conditional test.

(?(?=regex)then|else)

Regular Expression Conditionals

answered Oct 29 '08 at 15:56

Josh

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score 2 · Answer 7 · answered Oct 29 '08 at 17:59

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)

score 1 · Answer 8 · edited Jul 06 '15 at 03:39

1

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

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

edited Jul 06 '15 at 03:39

Undo

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answered Jul 06 '15 at 03:19

qmckinsey

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score 0 · Answer 9 · answered Oct 29 '08 at 15:15

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/

score 0 · Answer 10 · answered Oct 29 '08 at 15:23

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...

score 0 · Answer 11 · answered Oct 29 '08 at 15:27

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.

score 0 · Answer 12 · answered Oct 29 '08 at 15:41

0

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

And use only the first capture group.

answered Oct 29 '08 at 15:41

Andrew Cowenhoven

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score 0 · Answer 13 · answered Oct 29 '08 at 16:44

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"

Exclusive Or in Regular Expression

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