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A Regex that will never be matched by anything
(30 answers)
Closed 5 years ago.
I am studying regex lookaround from this tutorial.
It has an example explaining how lookaround are used to check the existence (or non-existence), but the regex inside lookaround parentheses is not used for actual match.
In example, the pattern q(?=u)i is checked against String quit. And it doesn't return a match.
I understood the example.
But I can't think of any string that matches this regex pattern. If my understanding of lookaround is correct I think there isn't any String that matches with this regex.
Am I correct? If not, which String matches this regex?
I'm not a big fan of this tutorial site, but if you read closely what it actually says, it never claims that the regex q(?=u)i matches the string quit:
Let's take one more look inside, to make sure you understand the implications of the lookahead. Let's apply q(?=u)i to quit. The lookahead is now positive and is followed by another token. Again, q matches q and u matches u. Again, the match from the lookahead must be discarded, so the engine steps back from i in the string to u. The lookahead was successful, so the engine continues with i. But i cannot match u. So this match attempt fails. All remaining attempts fail as well, because there are no more q's in the string.
I think you might still be confused about how lookaheads work. Either that, or you misread the tutorial site. If the former, then lookaheads work by asserting a match, without actually consuming anything in the string. So the regex q(?=u)i says to:
match the letter 'q'
lookahead to the next character after 'q' and assert that it is 'u'
then match an 'i' immediately after the 'q'
Of course, the string 'quit' fails, and in fact all strings would fail. The lookahead says to verify that q is followed by u, but the following pattern contradicts this by insisting that i is what follows.
Related
(Note: not a duplicate of Why can't you use repetition quantifiers in zero-width look behind assertions; see end of post.)
I'm trying to write a grep -P (Perl) regex that matches B, when it is not preceded by A -- regardless of whether there is intervening whitespace.
So, I tried this negative lookbehind, and tested it in regex101.com:
(?<!A)\s*B
This causes "AB" not to be matched, which is good, but "A B" does result in a match, which is not what I want.
I am not exactly sure why this is. It has something to do with the fact that \s* matches the empty string "", and you can say that there are, as such, infinity matches of \s* between A and B. But why does this affect "A B" but not "AB"?
Is the following regex a proper solution, and if so, why exactly does it fix the problem?
(?<![A\s])\s*B
I posted this before and it was incorrectly marked as a duplicate question. The variable-length thing I'm looking for is part of the match, not part of the negative lookbehind itself -- so this quite different from the other question. Yes, I could put the \s* inside the negative lookbehind, but I haven't done so (and doing so is not supported, as the other question explains). Also, I am particularly interested in why the alternate regex I post above works, since I know it works but I'm not exactly sure why. The other question did not help answer that.
But why does this affect "A B" but not "AB"?
Regexes match at a position, which it is helpful to think of as being between characters. In "A B" there is a position (after the space and before the B) where (?<!A) succeeds (because there isn't an A immediately preceding; there's a space instead), and \s*B succeeds (\s* matches the empty string, and B matches B), so the entire pattern succeeds.
In "AB" there is no such position. The only place where \s*B can match (immediately before the B), is also immediately after the A, so (?<!A) cannot succeed. There are no positions that satisfy both, so the pattern as a whole can't succeed.
Is the following regex a proper solution, and if so, why exactly does it fix the problem?
(?<![A\s])\s*B
This works because (?<![A\s]) will not succeed immediately after an A or after a space. So now the lookbehind forbids any match position that has spaces before it. If there are any spaces before the B, they have to be consumed by the \s* portion of the pattern, and the match position must be before them. If that position also doesn't have an A before it, the lookbehind can succeed and the pattern as a whole can match.
This is a trick that's made possible by the fact that \s is a fixed-width pattern that matches at every position inside of a non-empty \s* match. It can't be extended to the general case of any pattern between the (non-)A and the B.
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How to negate specific word in regex? [duplicate]
(12 answers)
Closed 7 years ago.
I am trying to learn RegEx and build a regular expression that would look whether specified word is NOT in the provided string. So far I did try Regular Expression Info and RexxEgg all this tested on Regular Expression Online but I did not find the answer to my question.
I have tried conditionals and lookarounds. Let's say I want to build an expression to test against not existing word myword and pass expression when the word is NOT in the string. I used expression
(?(?!myword).*)
but RegEx passes regardless the word myword meaning both strings This is the text and This is myword the text pass the test.
Using negative lookahead and conditions is used to test that condition is true when myword does not exist. Lookahead is also zero length and therefore .* would return the whole string.
Hope someone can help :)
^(?(?!\bmyword\b).)*$
You can try this.See demo.Also use \b for matching exactly myword and not mywords
https://regex101.com/r/hI0qP0/7
You should use anchors and negative lookahead:
^(?!.*?myword).*$
(?!.*?myword) is a negative lookahead that will fail the match if myword is found anywhere in the input string.
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Reference - What does this regex mean?
(1 answer)
Closed 8 years ago.
I have the following regex :
.*(?:(?:(?<!a)cc|string).*number).*
And I am trying to understand what the ? in the beginning of the string between brackets mean. I know the a? means that the previous character 'a' can be repeated zero or one time. But what does it mean when it appears in the beginning of a string ?
The answer requires a little history lesson. When Larry Wall wanted to add new features to regexes in Perl, he couldn't just change the meaning of existing metacharacters, or assign special meanings to characters that didn't have them. That would have broken a lot of regexes that had been working. Instead, he had to look for character sequences that would never appear in a regex.
There was only the one kind of group originally: what we now call capturing groups. The opening parenthesis was a metacharacter, so it would make no sense to follow it with a quantifier. You could match a literal open-paren zero or one time with \(?, or you could match (and capture) a literal question mark with (\?), but if you tried to use (? in regex it would throw an exception.
Larry changed the rule so (? could appear in a regex, but it must form the beginning of a special-group construct, which requires at least one more character. So, to answer your question, the string doesn't start with ?. The sequence (?: forms a single token, representing the beginning of a non-capturing group. We also have (?= and (?! for positive and negative lookaheads, (?<= and (?<! for lookbehinds, and so on.
(?:) is a non-capturing group. It do a matching operation only. It won't capture anything.
(?<!) is a Negative lookbehind.
from regular-expressions.info (emphasis added)
Let's take one more look inside, to make sure you understand the
implications of the lookahead. Let's apply q(?=u)i to quit. The
lookahead is now positive and is followed by another token. Again, q
matches q and u matches u. Again, the match from the lookahead must be
discarded, so the engine steps back from i in the string to u. The
lookahead was successful, so the engine continues with i. But i cannot
match u. So this match attempt fails. All remaining attempts fail as
well, because there are no more q's in the string.
does this necessarily mean that the match has to stop after this q not followed by a u is matched? What can come after once the q is matched? What if we want to perform more matches after this q not followed by a u? eg, if I want to continue to match the rest of the letters in the word quote? q(?=u)ote.
Yes, once the lookahead assertion fails then the match stops. In that sense they're no different from any other part of a regex - if they don't "match" then the overall match fails. The difference is that the matching characters are not consumed by the match, so the following part of the regex (if there is one) still needs to match those characters.
In this case the u matches the lookahead (?=u), but the u isn't consumed by the match. Therefore in the next step the u is tested against the i and the overall match fails. Using q(?=u) means the q must be followed by a u. To match quit using a similar regex you could use q(?=u)uit.
If you want to match after a q not followed by a u then you could use a negative lookahead instead of a positive lookahead, e.g. q(?!u)ote would match qote, but these examples are contrived. Lookaheads (and Lookbehinds) are very useful but they take some getting used to, and they're not needed in the vast majority of cases.
I have been solving old question from stack so that I can improve my regex knowledge. As I have a basic knowledge of regex, most of them were easy but this question regex problem is tough.
It asks for a regex that extracts from this kind of string ou=persons,ou=(.*),dc=company,dc=org the last string immediately preceded by a comma not followed by (.*). In the last case, this should give dc=company,dc=org.
The solution is (?<=,(?!.*\Q(.*)\E)).* but I cannot understand its flow. I understood (?!.*\Q(.*)\E) portion but other are still mystery to me. Specially ?<= which is a positive look-behind. Does it search from end of string? Can anyone explain it to me like I am a 7 year old kid — and please http://regex101.com/ is not helping.
The RegEx (?<=,(?!.*\Q(.*)\E)).* look-behind potion works like this:
Start at the beginning of the string at first character.
Can we match the the thing we are looking for? ,(?!.*\Q(.*)\E)
If we can't: Move forward one character, Go To 2. and check match again.
If a match is found: Capture all the remaining characters until we can't find any .* (or generally then try the matching the remaining RegEx).
For a more wordly explaination consider reading Lookahead and Lookbehind Zero-Length Assertions.
A lookbehind allows you to specify a context just before the actual match.
You can say ,(dc=) and only return the capture group, or ,\Kdc=, or (?<=,)dc= to return the match on dc= but require that the comma is present just before the match.
The facility also allows for multiple lookbehinds, so you could do (?<=a.*)(?<=b.*)c to match c only if it is preceded by both a and b somewhere in the input.
A lookbehind is basically syntactic sugar, in that you can usually rephrase your conditions using some other regex construct. It can be really handy when you have multiple unanchored constraints, like in the last example