CODEWITHSUNDEEP
regexxor
SocialCensus
SocialCensus

Reputation: 1269

Exclusive Or in Regular Expression

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!

Upvotes: 39

Views: 63459

Answers (14)

Bion Alex Howard
Bion Alex Howard

Reputation: 104

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:

direct xor approach

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:

xor apple banana orange

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.

exclude 2 or more, match one

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)):

apple xor banana via or and nand

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:

not microsoft, and apple or banana, and apple nand banana

Upvotes: -2

oriadam
oriadam

Reputation: 8559

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

Upvotes: 10

qmckinsey
qmckinsey

Reputation: 45

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

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

Upvotes: 2

Tristan
Tristan

Reputation: 499

This is what I use:

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

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

Upvotes: 45

cjm
cjm

Reputation: 62099

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)

Upvotes: 2

Dave Sherohman
Dave Sherohman

Reputation: 46187

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.

Upvotes: 3

bill_the_loser
bill_the_loser

Reputation: 195

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"

Upvotes: 0

erickson
erickson

Reputation: 269667

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

Upvotes: 20

Josh
Josh

Reputation: 18690

You might want to consider the ? conditional test.

(?(?=regex)then|else)

Regular Expression Conditionals

Upvotes: 2

Andrew Cowenhoven
Andrew Cowenhoven

Reputation: 2816

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

And use only the first capture group.

Upvotes: 0

PhiLho
PhiLho

Reputation: 41142

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.

Upvotes: 0

Can Berk G&#252;der
Can Berk G&#252;der

Reputation: 113310

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...

Upvotes: 0

Robert K
Robert K

Reputation: 30328

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/

Upvotes: 0

Ed Guiness
Ed Guiness

Reputation: 35267

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/)

Upvotes: 9

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