Collapse and Capture a Repeating Pattern in a Single Regex Expression - regex

I keep bumping into situations where I need to capture a number of tokens from a string and after countless tries I couldn't find a way to simplify the process.
So let's say the text is:
start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end
This example has 8 items inside, but say it could have between 3 and 10 items.
I'd ideally like something like this:
start:(?:(\w+)-?){3,10}:end nice and clean BUT it only captures the last match. see here
I usually use something like this in simple situations:
start:(\w+)-(\w+)-(\w+)-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?:end
3 groups mandatory and another 7 optional because of the max 10 limit, but this doesn't look 'nice' and it would be a pain to write and track if the max limit was 100 and the matches were more complex. demo
And the best I could do so far:
start:(\w+)-((?1))-((?1))-?((?1))?-?((?1))?-?((?1))?-?((?1))?-?((?1))?:end
shorter especially if the matches are complex but still long. demo
Anyone managed to make it work as a 1 regex-only solution without programming?
I'm mostly interested on how can this be done in PCRE but other flavors would be ok too.
Update:
The purpose is to validate a match and capture individual tokens inside match 0 by RegEx alone, without any OS/Software/Programming-Language limitation
Update 2 (bounty):
With #nhahtdh's help I got to the RegExp below by using \G:
(?:start:(?=(?:[\w]+(?:-|(?=:end))){3,10}:end)|(?!^)\G-)([\w]+)
demo even shorter, but can be described without repeating code
I'm also interested in the ECMA flavor and as it doesn't support \G wondering if there's another way, especially without using /g modifier.

Read this first!
This post is to show the possibility rather than endorsing the "everything regex" approach to problem. The author has written 3-4 variations, each has subtle bug that are tricky to detect, before reaching the current solution.
For your specific example, there are other better solution that is more maintainable, such as matching and splitting the match along the delimiters.
This post deals with your specific example. I really doubt a full generalization is possible, but the idea behind is reusable for similar cases.
Summary
.NET supports capturing repeating pattern with CaptureCollection class.
For languages that supports \G and look-behind, we may be able to construct a regex that works with global matching function. It is not easy to write it completely correct and easy to write a subtly buggy regex.
For languages without \G and look-behind support: it is possible to emulate \G with ^, by chomping the input string after a single match. (Not covered in this answer).
Solution
This solution assumes the regex engine supports \G match boundary, look-ahead (?=pattern), and look-behind (?<=pattern). Java, Perl, PCRE, .NET, Ruby regex flavors support all those advanced features above.
However, you can go with your regex in .NET. Since .NET supports capturing all instances of that is matched by a capturing group that is repeated via CaptureCollection class.
For your case, it can be done in one regex, with the use of \G match boundary, and look-ahead to constrain the number of repetitions:
(?:start:(?=\w+(?:-\w+){2,9}:end)|(?<=-)\G)(\w+)(?:-|:end)
DEMO. The construction is \w+- repeated, then \w+:end.
(?:start:(?=\w+(?:-\w+){2,9}:end)|(?!^)\G-)(\w+)
DEMO. The construction is \w+ for the first item, then -\w+ repeated. (Thanks to ka ᵠ for the suggestion). This construction is simpler to reason about its correctness, since there are less alternations.
\G match boundary is especially useful when you need to do tokenization, where you need to make sure the engine not skipping ahead and matching stuffs that should have been invalid.
Explanation
Let us break down the regex:
(?:
start:(?=\w+(?:-\w+){2,9}:end)
|
(?<=-)\G
)
(\w+)
(?:-|:end)
The easiest part to recognize is (\w+) in the line before last, which is the word that you want to capture.
The last line is also quite easy to recognize: the word to be matched may be followed by - or :end.
I allow the regex to freely start matching anywhere in the string. In other words, start:...:end can appear anywhere in the string, and any number of times; the regex will simply match all the words. You only need to process the array returned to separate where the matched tokens actually come from.
As for the explanation, the beginning of the regex checks for the presence of the string start:, and the following look-ahead checks that the number of words is within specified limit and it ends with :end. Either that, or we check that the character before the previous match is a -, and continue from previous match.
For the other construction:
(?:
start:(?=\w+(?:-\w+){2,9}:end)
|
(?!^)\G-
)
(\w+)
Everything is almost the same, except that we match start:\w+ first before matching the repetition of the form -\w+. In contrast to the first construction, where we match start:\w+- first, and the repeated instances of \w+- (or \w+:end for the last repetition).
It is quite tricky to make this regex works for matching in middle of the string:
We need to check the number of words between start: and :end (as part of the requirement of the original regex).
\G matches the beginning of the string also! (?!^) is needed to prevent this behavior. Without taking care of this, the regex may produce a match when there isn't any start:.
For the first construction, the look-behind (?<=-) already prevent this case ((?!^) is implied by (?<=-)).
For the first construction (?:start:(?=\w+(?:-\w+){2,9}:end)|(?<=-)\G)(\w+)(?:-|:end), we need to make sure that we don't match anything funny after :end. The look-behind is for that purpose: it prevents any garbage after :end from matching.
The second construction doesn't run into this problem, since we will get stuck at : (of :end) after we have matched all the tokens in between.
Validation Version
If you want to do validation that the input string follows the format (no extra stuff in front and behind), and extract the data, you can add anchors as such:
(?:^start:(?=\w+(?:-\w+){2,9}:end$)|(?!^)\G-)(\w+)
(?:^start:(?=\w+(?:-\w+){2,9}:end$)|(?!^)\G)(\w+)(?:-|:end)
(Look-behind is also not needed, but we still need (?!^) to prevent \G from matching the start of the string).
Construction
For all the problems where you want to capture all instances of a repetition, I don't think there exists a general way to modify the regex. One example of a "hard" (or impossible?) case to convert is when a repetition has to backtrack one or more loop to fulfill certain condition to match.
When the original regex describes the whole input string (validation type), it is usually easier to convert compared to a regex that tries to match from the middle of the string (matching type). However, you can always do a match with the original regex, and we convert matching type problem back to validation type problem.
We build such regex by going through these steps:
Write a regex that covers the part before the repetition (e.g. start:). Let us call this prefix regex.
Match and capture the first instance. (e.g. (\w+))
(At this point, the first instance and delimiter should have been matched)
Add the \G as an alternation. Usually also need to prevent it from matching the start of the string.
Add the delimiter (if any). (e.g. -)
(After this step, the rest of the tokens should have also been matched, except the last maybe)
Add the part that covers the part after the repetition (if necessary) (e.g. :end). Let us call the part after the repetition suffix regex (whether we add it to the construction doesn't matter).
Now the hard part. You need to check that:
There is no other way to start a match, apart from the prefix regex. Take note of the \G branch.
There is no way to start any match after the suffix regex has been matched. Take note of how \G branch starts a match.
For the first construction, if you mix the suffix regex (e.g. :end) with delimiter (e.g. -) in an alternation, make sure you don't end up allowing the suffix regex as delimiter.

Although it might theoretically be possible to write a single expression, it's a lot more practical to match the outer boundaries first and then perform a split on the inner part.
In ECMAScript I would write it like this:
'start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end'
.match(/^start:([\w-]+):end$/)[1] // match the inner part
.split('-') // split inner part (this could be a split regex as well)
In PHP:
$txt = 'start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end';
if (preg_match('/^start:([\w-]+):end$/', $txt, $matches)) {
print_r(explode('-', $matches[1]));
}

Of course you can use the regex in this quoted string.
"(?<a>\\w+)-(?<b>\\w+)-(?:(?<c>\\w+)" \
"(?:-(?<d>\\w+)(?:-(?<e>\\w+)(?:-(?<f>\\w+)" \
"(?:-(?<g>\\w+)(?:-(?<h>\\w+)(?:-(?<i>\\w+)" \
"(?:-(?<j>\\w+))?" \
")?)?)?" \
")?)?)?" \
")"
Is it a good idea? No, I don't think so.

Not sure you can do it in that way, but you can use the global flag to find all of the words between the colons, see:
http://regex101.com/r/gK0lX1
You'd have to validate the number of groups yourself though. Without the global flag you're only getting a single match, not all matches - change {3,10} to {1,5} and you get the result 'sir' instead.
import re
s = "start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end"
print re.findall(r"(\b\w+?\b)(?:-|:end)", s)
produces
['test', 'test', 'lorem', 'ipsum', 'sir', 'doloret', 'etc', 'etc', 'something']

When you combine:
Your observation: any kind of repitition of a single capture group will result in an overwrite of the last capture, thus returning only the last capture of the capture group.
The knowledge: Any kind of capturing based on the parts, instead of the whole, makes it impossible to set a limit on the amount of times the regex engine will repeat. The limit would have to be metadata (not regex).
With a requirement that the answer cannot involve programming (looping), nor an answer that involves simply copy-pasting capturegroups as you've done in your question.
It can be deduced that it cannot be done.
Update: There are some regex engines for which p. 1 is not necessarily true. In that case the regex you have indicated start:(?:(\w+)-?){3,10}:end will do the job (source).

Related

How to write a regular expression inside awk to IGNORE a word as a whole? [duplicate]

I've been looking around and could not make this happen. I am not totally noob.
I need to get text delimited by (including) START and END that doesn't contain START. Basically I can't find a way to negate a whole word without using advanced stuff.
Example string:
abcSTARTabcSTARTabcENDabc
The expected result:
STARTabcEND
Not good:
STARTabcSTARTabcEND
I can't use backward search stuff. I am testing my regex here: www.regextester.com
Thanks for any advice.
Try this
START(?!.*START).*?END
See it here online on Regexr
(?!.*START) is a negative lookahead. It ensures that the word "START" is not following
.*? is a non greedy match of all characters till the next "END". Its needed, because the negative lookahead is just looking ahead and not capturing anything (zero length assertion)
Update:
I thought a bit more, the solution above is matching till the first "END". If this is not wanted (because you are excluding START from the content) then use the greedy version
START(?!.*START).*END
this will match till the last "END".
START(?:(?!START).)*END
will work with any number of START...END pairs. To demonstrate in Python:
>>> import re
>>> a = "abcSTARTdefENDghiSTARTjlkENDopqSTARTrstSTARTuvwENDxyz"
>>> re.findall(r"START(?:(?!START).)*END", a)
['STARTdefEND', 'STARTjlkEND', 'STARTuvwEND']
If you only care for the content between START and END, use this:
(?<=START)(?:(?!START).)*(?=END)
See it here:
>>> re.findall(r"(?<=START)(?:(?!START).)*(?=END)", a)
['def', 'jlk', 'uvw']
The really pedestrian solution would be START(([^S]|S*S[^ST]|ST[^A]|STA[^R]|STAR[^T])*(S(T(AR?)?)?)?)END. Modern regex flavors have negative assertions which do this more elegantly, but I interpret your comment about "backwards search" to perhaps mean you cannot or don't want to use this feature.
Update: Just for completeness, note that the above is greedy with respect to the end delimiter. To only capture the shortest possible string, extend the negation to also cover the end delimiter -- START(([^ES]|E*E[^ENS]|EN[^DS]|S*S[^STE]|ST[^AE]|STA[^RE]|STAR[^TE])*(S(T(AR?)?)?|EN?)?)END. This risks to exceed the torture threshold in most cultures, though.
Bug fix: A previous version of this answer had a bug, in that SSTART could be part of the match (the second S would match [^T], etc). I fixed this but by the addition of S in [^ST] and adding S* before the non-optional S to allow for arbitrary repetitions of S otherwise.
May I suggest a possible improvement on the solution of Tim Pietzcker?
It seems to me that START(?:(?!START).)*?END is better in order to only catch a START immediately followed by an END without any START or END in between. I am using .NET and Tim's solution would match also something like START END END. At least in my personal case this is not wanted.
[EDIT: I have left this post for the information on capture groups but the main solution I gave was not correct.
(?:START)((?:[^S]|S[^T]|ST[^A]|STA[^R]|STAR[^T])*)(?:END)
as pointed out in the comments would not work; I was forgetting that the ignored characters could not be dropped and thus you would need something such as ...|STA(?![^R])| to still allow that character to be part of END, thus failing on something such as STARTSTAEND; so it's clearly a better choice; the following should show the proper way to use the capture groups...]
The answer given using the 'zero-width negative lookahead' operator "?!", with capture groups, is: (?:START)((?!.*START).*)(?:END) which captures the inner text using $1 for the replace. If you want to have the START and END tags captured you could do (START)((?!.*START).*)(END) which gives $1=START $2=text and $3=END or various other permutations by adding/removing ()s or ?:s.
That way if you are using it to do search and replace, you can do, something like BEGIN$1FINISH. So, if you started with:
abcSTARTdefSTARTghiENDjkl
you would get ghi as capture group 1, and replacing with BEGIN$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.

Can this be parsed by regular expression [duplicate]

I keep bumping into situations where I need to capture a number of tokens from a string and after countless tries I couldn't find a way to simplify the process.
So let's say the text is:
start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end
This example has 8 items inside, but say it could have between 3 and 10 items.
I'd ideally like something like this:
start:(?:(\w+)-?){3,10}:end nice and clean BUT it only captures the last match. see here
I usually use something like this in simple situations:
start:(\w+)-(\w+)-(\w+)-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?-?(\w+)?:end
3 groups mandatory and another 7 optional because of the max 10 limit, but this doesn't look 'nice' and it would be a pain to write and track if the max limit was 100 and the matches were more complex. demo
And the best I could do so far:
start:(\w+)-((?1))-((?1))-?((?1))?-?((?1))?-?((?1))?-?((?1))?-?((?1))?:end
shorter especially if the matches are complex but still long. demo
Anyone managed to make it work as a 1 regex-only solution without programming?
I'm mostly interested on how can this be done in PCRE but other flavors would be ok too.
Update:
The purpose is to validate a match and capture individual tokens inside match 0 by RegEx alone, without any OS/Software/Programming-Language limitation
Update 2 (bounty):
With #nhahtdh's help I got to the RegExp below by using \G:
(?:start:(?=(?:[\w]+(?:-|(?=:end))){3,10}:end)|(?!^)\G-)([\w]+)
demo even shorter, but can be described without repeating code
I'm also interested in the ECMA flavor and as it doesn't support \G wondering if there's another way, especially without using /g modifier.
Read this first!
This post is to show the possibility rather than endorsing the "everything regex" approach to problem. The author has written 3-4 variations, each has subtle bug that are tricky to detect, before reaching the current solution.
For your specific example, there are other better solution that is more maintainable, such as matching and splitting the match along the delimiters.
This post deals with your specific example. I really doubt a full generalization is possible, but the idea behind is reusable for similar cases.
Summary
.NET supports capturing repeating pattern with CaptureCollection class.
For languages that supports \G and look-behind, we may be able to construct a regex that works with global matching function. It is not easy to write it completely correct and easy to write a subtly buggy regex.
For languages without \G and look-behind support: it is possible to emulate \G with ^, by chomping the input string after a single match. (Not covered in this answer).
Solution
This solution assumes the regex engine supports \G match boundary, look-ahead (?=pattern), and look-behind (?<=pattern). Java, Perl, PCRE, .NET, Ruby regex flavors support all those advanced features above.
However, you can go with your regex in .NET. Since .NET supports capturing all instances of that is matched by a capturing group that is repeated via CaptureCollection class.
For your case, it can be done in one regex, with the use of \G match boundary, and look-ahead to constrain the number of repetitions:
(?:start:(?=\w+(?:-\w+){2,9}:end)|(?<=-)\G)(\w+)(?:-|:end)
DEMO. The construction is \w+- repeated, then \w+:end.
(?:start:(?=\w+(?:-\w+){2,9}:end)|(?!^)\G-)(\w+)
DEMO. The construction is \w+ for the first item, then -\w+ repeated. (Thanks to ka ᵠ for the suggestion). This construction is simpler to reason about its correctness, since there are less alternations.
\G match boundary is especially useful when you need to do tokenization, where you need to make sure the engine not skipping ahead and matching stuffs that should have been invalid.
Explanation
Let us break down the regex:
(?:
start:(?=\w+(?:-\w+){2,9}:end)
|
(?<=-)\G
)
(\w+)
(?:-|:end)
The easiest part to recognize is (\w+) in the line before last, which is the word that you want to capture.
The last line is also quite easy to recognize: the word to be matched may be followed by - or :end.
I allow the regex to freely start matching anywhere in the string. In other words, start:...:end can appear anywhere in the string, and any number of times; the regex will simply match all the words. You only need to process the array returned to separate where the matched tokens actually come from.
As for the explanation, the beginning of the regex checks for the presence of the string start:, and the following look-ahead checks that the number of words is within specified limit and it ends with :end. Either that, or we check that the character before the previous match is a -, and continue from previous match.
For the other construction:
(?:
start:(?=\w+(?:-\w+){2,9}:end)
|
(?!^)\G-
)
(\w+)
Everything is almost the same, except that we match start:\w+ first before matching the repetition of the form -\w+. In contrast to the first construction, where we match start:\w+- first, and the repeated instances of \w+- (or \w+:end for the last repetition).
It is quite tricky to make this regex works for matching in middle of the string:
We need to check the number of words between start: and :end (as part of the requirement of the original regex).
\G matches the beginning of the string also! (?!^) is needed to prevent this behavior. Without taking care of this, the regex may produce a match when there isn't any start:.
For the first construction, the look-behind (?<=-) already prevent this case ((?!^) is implied by (?<=-)).
For the first construction (?:start:(?=\w+(?:-\w+){2,9}:end)|(?<=-)\G)(\w+)(?:-|:end), we need to make sure that we don't match anything funny after :end. The look-behind is for that purpose: it prevents any garbage after :end from matching.
The second construction doesn't run into this problem, since we will get stuck at : (of :end) after we have matched all the tokens in between.
Validation Version
If you want to do validation that the input string follows the format (no extra stuff in front and behind), and extract the data, you can add anchors as such:
(?:^start:(?=\w+(?:-\w+){2,9}:end$)|(?!^)\G-)(\w+)
(?:^start:(?=\w+(?:-\w+){2,9}:end$)|(?!^)\G)(\w+)(?:-|:end)
(Look-behind is also not needed, but we still need (?!^) to prevent \G from matching the start of the string).
Construction
For all the problems where you want to capture all instances of a repetition, I don't think there exists a general way to modify the regex. One example of a "hard" (or impossible?) case to convert is when a repetition has to backtrack one or more loop to fulfill certain condition to match.
When the original regex describes the whole input string (validation type), it is usually easier to convert compared to a regex that tries to match from the middle of the string (matching type). However, you can always do a match with the original regex, and we convert matching type problem back to validation type problem.
We build such regex by going through these steps:
Write a regex that covers the part before the repetition (e.g. start:). Let us call this prefix regex.
Match and capture the first instance. (e.g. (\w+))
(At this point, the first instance and delimiter should have been matched)
Add the \G as an alternation. Usually also need to prevent it from matching the start of the string.
Add the delimiter (if any). (e.g. -)
(After this step, the rest of the tokens should have also been matched, except the last maybe)
Add the part that covers the part after the repetition (if necessary) (e.g. :end). Let us call the part after the repetition suffix regex (whether we add it to the construction doesn't matter).
Now the hard part. You need to check that:
There is no other way to start a match, apart from the prefix regex. Take note of the \G branch.
There is no way to start any match after the suffix regex has been matched. Take note of how \G branch starts a match.
For the first construction, if you mix the suffix regex (e.g. :end) with delimiter (e.g. -) in an alternation, make sure you don't end up allowing the suffix regex as delimiter.
Although it might theoretically be possible to write a single expression, it's a lot more practical to match the outer boundaries first and then perform a split on the inner part.
In ECMAScript I would write it like this:
'start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end'
.match(/^start:([\w-]+):end$/)[1] // match the inner part
.split('-') // split inner part (this could be a split regex as well)
In PHP:
$txt = 'start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end';
if (preg_match('/^start:([\w-]+):end$/', $txt, $matches)) {
print_r(explode('-', $matches[1]));
}
Of course you can use the regex in this quoted string.
"(?<a>\\w+)-(?<b>\\w+)-(?:(?<c>\\w+)" \
"(?:-(?<d>\\w+)(?:-(?<e>\\w+)(?:-(?<f>\\w+)" \
"(?:-(?<g>\\w+)(?:-(?<h>\\w+)(?:-(?<i>\\w+)" \
"(?:-(?<j>\\w+))?" \
")?)?)?" \
")?)?)?" \
")"
Is it a good idea? No, I don't think so.
Not sure you can do it in that way, but you can use the global flag to find all of the words between the colons, see:
http://regex101.com/r/gK0lX1
You'd have to validate the number of groups yourself though. Without the global flag you're only getting a single match, not all matches - change {3,10} to {1,5} and you get the result 'sir' instead.
import re
s = "start:test-test-lorem-ipsum-sir-doloret-etc-etc-something:end"
print re.findall(r"(\b\w+?\b)(?:-|:end)", s)
produces
['test', 'test', 'lorem', 'ipsum', 'sir', 'doloret', 'etc', 'etc', 'something']
When you combine:
Your observation: any kind of repitition of a single capture group will result in an overwrite of the last capture, thus returning only the last capture of the capture group.
The knowledge: Any kind of capturing based on the parts, instead of the whole, makes it impossible to set a limit on the amount of times the regex engine will repeat. The limit would have to be metadata (not regex).
With a requirement that the answer cannot involve programming (looping), nor an answer that involves simply copy-pasting capturegroups as you've done in your question.
It can be deduced that it cannot be done.
Update: There are some regex engines for which p. 1 is not necessarily true. In that case the regex you have indicated start:(?:(\w+)-?){3,10}:end will do the job (source).

Regex to find last occurrence of pattern in a string

My string being of the form:
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
I only want to match against the last segment of whitespace before the last period(.)
So far I am able to capture whitespace but not the very last occurrence using:
\s+(?=\.\w)
How can I make it less greedy?
In a general case, you can match the last occurrence of any pattern using the following scheme:
pattern(?![\s\S]*pattern)
(?s)pattern(?!.*pattern)
pattern(?!(?s:.*)pattern)
where [\s\S]* matches any zero or more chars as many as possible. (?s) and (?s:.) can be used with regex engines that support these constructs so as to use . to match any chars.
In this case, rather than \s+(?![\s\S]*\s), you may use
\s+(?!\S*\s)
See the regex demo. Note the \s and \S are inverse classes, thus, it makes no sense using [\s\S]* here, \S* is enough.
Details:
\s+ - one or more whitespace chars
(?!\S*\s) - that are not immediately followed with any 0 or more non-whitespace chars and then a whitespace.
You can try like so:
(\s+)(?=\.[^.]+$)
(?=\.[^.]+$) Positive look ahead for a dot and characters except dot at the end of line.
Demo:
https://regex101.com/r/k9VwC6/3
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=((?<=\S)\s+)).*
replaced by `>\1<`
> <
As a more generalized example
This example defines several needles and finds the last occurrence of either one of them. In this example the needles are:
defined word findMyLastOccurrence
whitespaces (?<=\S)\s+
dots (?<=[^\.])\.+
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)).*
replaced by `>\1<`
>..<
Explanation:
Part 1 .*
is greedy and finds everything as long as the needles are found. Thus, it also captures all needle occurrences until the very last needle.
edit to add:
in case we are interested in the first hit, we can prevent the greediness by writing .*?
Part 2 (?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+|(?<=**Not**NeedlePart)NeedlePart+))
defines the 'break' condition for the greedy 'find-all'. It consists of several parts:
(?=(needles))
positive lookahead: ensure that previously found everything is followed by the needles
findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)|(?<=**Not**NeedlePart)NeedlePart+
several needles for which we are looking. Needles are patterns themselves.
In case we look for a collection of whitespaces, dots or other needleparts, the pattern we are looking for is actually: anything which is not a needlepart, followed by one or more needleparts (thus needlepart is +). See the example for whitespaces \s negated with \S, actual dot . negated with [^.]
Part 3 .*
as we aren't interested in the remainder, we capture it and dont use it any further. We could capture it with parenthesis and use it as another group, but that's out of scope here
SIMPLE SOLUTION for a COMMON PROBLEM
All of the answers that I have read through are way off topic, overly complicated, or just simply incorrect. This question is a common problem that regex offers a simple solution for.
Breaking Down the General Problem
THE STRING
The generalized problem is such that there is a string that contains several characters.
THE SUB-STRING
Within the string is a sub-string made up of a few characters. Often times this is a file extension (i.e .c, .ts, or .json), or a top level domain (i.e. .com, .org, or .io), but it could be something as arbitrary as MC Donald's Mulan Szechuan Sauce. The point it is, it may not always be something simple.
THE BEFORE VARIANCE (Most important part)
The before variance is an arbitrary character, or characters, that always comes just before the sub-string. In this question, the before variance is an unknown amount of white-space. Its a variance because the amount of white-space that needs to be match against varies (or has a dynamic quantity).
Describing the Solution in Reference to the Problem
(Solution Part 1)
Often times when working with regular expressions its necessary to work in reverse.
We will start at the end of the problem described above, and work backwards, henceforth; we are going to start at the The Before Variance (or #3)
So, as mentioned above, The Before Variance is an unknown amount of white-space. We know that it includes white-space, but we don't know how much, so we will use the meta sequence for Any Whitespce with the one or more quantifier.
The Meta Sequence for "Any Whitespace" is \s.
The "One or More" quantifier is +
so we will start with...
NOTE: In ECMAS Regex the / characters are like quotes around a string.
const regex = /\s+/g
I also included the g to tell the engine to set the global flag to true. I won't explain flags, for the sake of brevity, but if you don't know what the global flag does, you should DuckDuckGo it.
(Solution Part 2)
Remember, we are working in reverse, so the next part to focus on is the Sub-string. In this question it is .com, but the author may want it to match against a value with variance, rather than just the static string of characters .com, therefore I will talk about that more below, but to stay focused, we will work with .com for now.
It's necessary that we use a concept here that's called ZERO LENGTH ASSERTION. We need a "zero-length assertion" because we have a sub-string that is significant, but is not what we want to match against. "Zero-length assertions" allow us to move the point in the string where the regular expression engine is looking at, without having to match any characters to get there.
The Zero-Length Assertion that we are going to use is called LOOK AHEAD, and its syntax is as follows.
Look-ahead Syntax: (?=Your-SubStr-Here)
We are going to use the look ahead to match against a variance that comes before the pattern assigned to the look-ahead, which will be our sub-string. The result looks like this:
const regex = /\s+(?=\.com)/gi
I added the insensitive flag to tell the engine to not be concerned with the case of the letter, in other words; the regular expression /\s+(?=\.cOM)/gi
is the same as /\s+(?=\.Com)/gi, and both are the same as: /\s+(?=\.com)/gi &/or /\s+(?=.COM)/gi. Everyone of the "Just Listed" regular expressions are equivalent so long as the i flag is set.
That's it! The link HERE (REGEX101) will take you to an example where you can play with the regular expression if you like.
I mentioned above working with a sub-string that has more variance than .com.
You could use (\s*)(?=\.\w{3,}) for instance.
The problem with this regex, is even though it matches .txt, .org, .json, and .unclepetespurplebeet, the regex isn't safe. When using the question's string of...
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
as an example, you can see at the LINK HERE (Regex101) there are 3 lines in the string. Those lines represent areas where the sub-string's lookahead's assertion returned true. Each time the assertion was true, a possibility for an incorrect final match was created. Though, only one match was returned in the end, and it was the correct match, when implemented in a program, or website, that's running in production, you can pretty much guarantee that the regex is not only going to fail, but its going to fail horribly and you will come to hate it.
You can try this. It will capture the last white space segment - in the first capture group.
(\s+)\.[^\.]*$

Multiple spaces, multiple commas and multiple hypens in alphanumeric regex

I am very new to regex and regular expressions, and I am stuck in a situation where I want to apply a regex on an JSF input field.
Where
alphanumeric
multiple spaces
multiple dot(.)
multiple hyphen (‐)
are allowed, and Minimum limit is 1 and Maximum limit is 5.
And for multiple values - they must be separated by comma (,)
So a Single value can be:
3kd-R
or
k3
or
-4
And multiple values (must be comma separated):
kdk30,3.K-4,ER--U,2,.I3,
By the help of stackoverflow, so far I am able to achieve only this:
(^[a-zA-Z0-9 ]{5}(,[a-zA-Z0-9 ]{5})*$)
Something like
^[-.a-zA-Z0-9 ]{1,5}(,[-.a-zA-Z0-9 ]{1,5})*$
Changes made
[-.a-zA-Z0-9 ] Added - and . to the character class so that those are matched as well.
{1,5} Quantifier, ensures that it is matched minimum 1 and maximum 5 characters
Regex demo
You've done pretty good. You need to add hyphen and dot to that first character class. Note: With the hyphen, since it delegates ranges within a character class, you need to position it where contextually it cannot be specifying a range--not to say put it where it seems like it would be an invalid range, e.g., 7-., but positionally cannot be a range, i.e., first or last. So your first character class would look something like this:
[a-zA-Z 0-9.-]{1,5} or [-a-zA-Z0-9 .]{1,5}
So, we've just defined what one segment looks like. That pattern can reoccur zero or more times. Of course, there are many ways to do that, but I would favor a regex subroutine because this allows code reuse. Now if the specs change or you're testing and realize you have to tweak that segment pattern, you only need to change it in one place.
Subroutines are not supported in BRE or ERE, but most widely-used modern regex engines support them (Perl, PCRE, Ruby, Delphi, R, PHP). They are very simple to use and understand. Basically, you just need to be able to refer to it (sound familiar? refer-back? back-reference?), so this means we need to capture the regex we wish to repeat. Then it's as simple as referring back to it, but instead of \1 which refers to the captured value (data), we want to refer to it as (?1), the capturing expression. In doing so, we've logically defined a subroutine:
([a-zA-Z 0-9.-]{1,5})(,(?1))*
So, the first group basically defines our subroutine and the second group consists of a comma followed by the same segment-definition expression we used for the first group, and that is optional ('*' is the zero-or-more quantifier).
If you operate on large quantities of data where efficiency is a consideration, don't capture when you don't have to. If your sole purpose for using parenthesis is to alternate (e.g., \b[bB](asset|eagle)\b hound) or to quantify, as in our second group, use the (?: ... ) notation, which signifies to the regex engine that this is a non-capturing group. Without going into great detail, there is a lot of overhead in maintaining the match locations--not that it's complex, per se, just potentially highly repetitive. Regex engines will match, store the information, then when the match fails, they "give up" the match and try again starting with the next matching substring. Each time they match your capture group, they're storing that information again. Okay, I'm off the soapbox now. :-)
So, we're almost there. I say "almost" because I don't have all the information. But if this should be the sole occupant of the "subject" (line, field, etc.--the data sample you're evaluating), you should anchor it to "assert" that requirement. The caret '^' is beginning of subject, and the dollar '$' is end of subject, so by encapsulating our expression in ^ ... $ we are asserting that the subject matches in it's entirety, front-to-back. These assertions have zero-length; they consume no data, only assert a relative position. You can operate on them, e.g., s/^/ / would indent your entire document two spaces. You haven't really substituted the beginning of line with two spaces, but you're able to operate on that imaginary, zero-length location. (Do some research on zero-length assertions [aka zero-width assertions, or look-arounds] to uncover a powerful feature of modern regex. For example, in the previous regex if I wanted to make sure I did not insert two spaces on blank lines: s/^(?!$)/ /)
Also, you didn't say if you need to capture the results to do something with it. My impression was it's validation only, so that's not necessary. However, if it is needed, you can wrap the entire expression in capturing parenthesis: ^( ... )$.
I'm going to provide a final solution that does not assume you need to capture but does assume the entire subject should consist of this value:
^([a-zA-Z 0-9. -]{1,5})(?:,(?1))*$
I know I went on a bit, but you said you were new to regex, so wanted to provide some detail. I hope it wasn't too much detail.
By the way, an excellent resource with tutorials is regular-expressions dot info, and a wonderful regex development and testing tool is regex101 dot com. And I can never say enough about stack overflow!

Look behinds: all the rage in regex?

Many regex questions lately have some kind of look-around element in the query that appears to me is not necessary to the success of the match. Is there some teaching resource that is promoting them? I am trying to figure out what kinds of cases you would be better off using a positive look ahead/behind. The main application I can see is when trying to not match an element. But, for example, this query from a recent question has a simple solution to capturing the .*, but why would you use a look behind?
(?<=<td><a href="\/xxx\.html\?n=[0-9]{0, 5}">).*(?=<\/a><span
And this one from another question:
$url = "www.example.com/id/1234";
preg_match("/\d+(?<=id\/[\d])/",$url,$matches);
When is it truly better to use a positive look-around? Can you give some examples?
I realize this is bordering on an opinion-based question, but I think the answers would be really instructive. Regex is confusing enough without making things more complicated... I have read this page and am more interested in some simple guidelines for when to use them rather than how they work.
Thanks for all the replies. In addition to those below, I recommend checking out m.buettner's great answer here.
You can capture overlapping matches, and you can find matches which could lie in the lookarounds of other matches.
You can express complex logical assertions about your match (because many engines let you use multiple lookbehind/lookahead assertions which all must match in order for the match to succeed).
Lookaround is a natural way to express the common constraint "matches X, if it is followed by/preceded by Y". It is (arguably) less natural to add extra "matching" parts that have to be thrown out by postprocessing.
Negative lookaround assertions, of course, are even more useful. Combined with #2, they can allow you do some pretty wizard tricks, which may even be hard to express in usual program logic.
Examples, by popular request:
Overlapping matches: suppose you want to find all candidate genes in a given genetic sequence. Genes generally start with ATG, and end with TAG, TAA or TGA. But, candidates could overlap: false starts may exist. So, you can use a regex like this:
ATG(?=((?:...)*(?:TAG|TAA|TGA)))
This simple regex looks for the ATG start-codon, followed by some number of codons, followed by a stop codon. It pulls out everything that looks like a gene (sans start codon), and properly outputs genes even if they overlap.
Zero-width matching: suppose you want to find every tr with a specific class in a computer-generated HTML page. You might do something like this:
<tr class="TableRow">.*?</tr>(?=<tr class="TableRow">|</table>)
This deals with the case in which a bare </tr> appears inside the row. (Of course, in general, an HTML parser is a better choice, but sometimes you just need something quick and dirty).
Multiple constraints: suppose you have a file with data like id:tag1,tag2,tag3,tag4, with tags in any order, and you want to find all rows with tags "green" and "egg". This can be done easily with two lookaheads:
(.*):(?=.*\bgreen\b)(?=.*\begg\b)
There are two great things about lookaround expressions:
They are zero-width assertions. They require to be matched, but they consume nothing of the input string. This allows to describe parts of the string which will not be contained in a match result. By using capturing groups in lookaround expressions, they are the only way to capture parts of the input multiple times.
They simplify a lot of things. While they do not extend regular languages, they easily allow to combine (intersect) multiple expressions to match the same part of a string.
Well one simple case where they are handy is when you are anchoring the pattern to the start or finish of a line, and just want to make sure that something is either right ahead or behind the pattern you are matching.
I try to address your points:
some kind of look-around element in the query that appears to me is not necessary to the success of the match
Of course they are necessary for the match. As soon as a lookaround assertions fails, there is no match. They can be used to ensure conditions around the pattern, that have additionally to be true. The whole regex does only match, if:
The pattern does fit and
The lookaround assertions are true.
==> But the returned match is only the pattern.
When is it truly better to use a positive look-around?
Simple answer: when you want stuff to be there, but you don't want to match it!
As Bergi mentioned in his answer, they are zero width assertions, this means they don't match a character sequence, they just ensure it is there. So the characters inside a lookaround expression are not "consumed", the regex engine continues after the last "consumed" character.
Regarding your first example:
(?<=<td><a href="\/xxx\.html\?n=[0-9]{0, 5}">).*(?=<\/a><span
I think there is a misunderstanding on your side, when you write "has a simple solution to capturing the .*". The .* is not "captured", it is the only thing that the expression does match. But only those characters are matched that have a "<td><a href="\/xxx\.html\?n=[0-9]{0, 5}">" before and a "<\/a><span" after (those two are not part of the match!).
"Captured" is only something that has been matched by a capturing group.
The second example
\d+(?<=id\/[\d])
Is interesting. It is matching a sequence of digits (\d+) and after the sequence, the lookbehind assertion checks if there is one digit with "id/" before it. Means it will fail if there is more than one digit or if the text "id/" before the digit is missing. Means this regex is matching only one digit, when there is fitting text before.
teaching resources
www.regular-expressions.info
perlretut on Looking ahead and looking behind
I'm assuming you understand the good uses of lookarounds, and ask why they are used with no apparent reason.
I think there are four main categories of how people use regular expressions:
Validation
Validation is usually done on the whole text. Lookarounds like you describe are not possible.
Match
Extracting a part of the text. Lookarounds are used mainly due to developer laziness: avoiding captures.
For example, if we have in a settings file with the line Index=5, we can match /^Index=(\d+)/ and take the first group, or match /(?<=^Index=)\d+/ and take everything.
As other answers said, sometimes you need overlapping between matches, but these are relatively rare.
Replace
This is similar to match with one difference: the whole match is removed and is being replaced with a new string (and some captured groups).
Example: we want to highlight the name in "Hi, my name is Bob!".
We can replace /(name is )(\w+)/ with $1<b>$2</b>,
but it is neater to replace /(?<=name is )\w+/ with <b>$&</b> - and no captures at all.
Split
split takes the text and breaks it to an array of tokens, with your pattern being the delimiter. This is done by:
Find a match. Everything before this match is token.
The content of the match is discarded, but:
In most flavors, each captured group in the match is also a token (notably not in Java).
When there are no more matches, the rest of the text is the last token.
Here, lookarounds are crucial. Matching a character means removing it from the result, or at least separating it from its token.
Example: We have a comma separated list of quoted string: "Hello","Hi, I'm Jim."
Splitting by comma /,/ is wrong: {"Hello", "Hi, I'm Jim."}
We can't add the quote mark, /",/: {"Hello, "Hi, I'm Jim."}
The only good option is lookbehind, /(?<="),/: {"Hello", "Hi, I'm Jim."}
Personally, I prefer to match the tokens rather than split by the delimiter, whenever that is possible.
Conclusion
To answer the main question - these lookarounds are used because:
Sometimes you can't match text that need.
Developers are shiftless.
Lookaround assertions can also be used to reduce backtracking which can be the main cause for a bad performance in regexes.
For example: The regex ^[0-9A-Z]([-.\w]*[0-9A-Z])*#(1) can also be written ^[0-9A-Z][-.\w]*(?<=[0-9A-Z])#(2) using a positive look behind (simple validation of the user name in an e-mail address).
Regex (1) can cause a lot of backtracking essentially because [0-9A-Z] is a subset of [-.\w] and the nested quantifiers. Regex (2) reduces the excessive backtracking, more information here Backtracking, section Controlling Backtracking > Lookbehind Assertions.
For more information about backtracking
Best Practices for Regular Expressions in the .NET Framework
Optimizing Regular Expression Performance, Part II: Taking Charge of Backtracking
Runaway Regular Expressions: Catastrophic Backtracking
I typed this a while back but got busy (still am, so I might take a while to reply back) and didn't get around to post it. If you're still open to answers...
Is there some teaching resource that is promoting them?
I don't think so, it's just a coincidence I believe.
But, for example, this query from a recent question has a simple solution to capturing the .*, but why would you use a look behind?
(?<=<td><a href="\/xxx\.html\?n=[0-9]{0, 5}">).*(?=<\/a><span
This is most probably a C# regex, since variable width lookbehinds are not supported my many regex engines. Well, the lookarounds could be certainly avoided here, because for this, I believe it's really simpler to have capture groups (and make the .* lazy as we're at it):
(<td><a href="\/xxx\.html\?n=[0-9]{0,5}">).*?(<\/a><span)
If it's for a replace, or
<td><a href="\/xxx\.html\?n=[0-9]{0,5}">(.*?)<\/a><span
for a match. Though an html parser would definitely be more advisable here.
Lookarounds in this case I believe are slower. See regex101 demo where the match is 64 steps for capture groups but 94+19 = 1-3 steps for the lookarounds.
When is it truly better to use a positive look-around? Can you give some examples?
Well, lookarounds have the property of being zero-width assertions, which mean they don't really comtribute to matches while they contribute onto deciding what to match and also allows overlapping matches.
Thinking a bit about it, I think, too, that negative lookarounds get used much more often, but that doesn't make positive lookarounds less useful!
Some 'exploits' I can find browsing some old answers of mine (links below will be demos from regex101) follow. When/If you see something you're not familiar about, I probably won't be explaining it here, since the question's focused on positive lookarounds, but you can always look at the demo links I provided where there's a description of the regex, and if you still want some explanation, let me know and I'll try to explain as much as I can.
To get matches between certain characters:
In some matches, positive lookahead make things easier, where a lookahead could do as well, or when it's not so practical to use no lookarounds:
Dog sighed. "I'm no super dog, nor special dog," said Dog, "I'm an ordinary dog, now leave me alone!" Dog pushed him away and made his way to the other dog.
We want to get all the dog (regardless of case) outside quotes. With a positive lookahead, we can do this:
\bdog\b(?=(?:[^"]*"[^"]*")*[^"]*$)
to ensure that there are even number of quotes ahead. With a negative lookahead, it would look like this:
\bdog\b(?!(?:[^"]*"[^"]*")*[^"]*"[^"]*$)
to ensure that there are no odd number of quotes ahead. Or use something like this if you don't want a lookahead, but you'll have to extract the group 1 matches:
(?:"[^"]+"[^"]+?)?(\bdog\b)
Okay, now say we want the opposite; find 'dog' inside the quotes. The regex with the lookarounds just need to have the sign inversed, first and second:
\bdog\b(?!(?:[^"]*"[^"]*")*[^"]*$)
\bdog\b(?=(?:[^"]*"[^"]*")*[^"]*"[^"]*$)
But without the lookaheads, it's not possible. the closest you can get is maybe this:
"[^"]*(\bdog\b)[^"]*"
But this doesn't get all the matches, or you can maybe use this:
"[^"]*?(\bdog\b)[^"]*?(?:(\bdog\b)[^"]*?)?"
But it's just not practical for more occurrences of dog and you get the results in variables with increasing numbers... And this is indeed easier with lookarounds, because they are zero width assertions, you don't have to worry about the expression inside the lookaround to match dog or not, or the regex wouldn't have obtained all the occurrences of dog in the quotes.
Of course now, this logic can be extended to groups of characters, such as getting specific patterns between words such as start and end.
Overlapping matches
If you have a string like:
abcdefghijkl
And want to extract all the consecutive 3 characters possible inside, you can use this:
(?=(...))
If you have something like:
1A Line1 Detail1 Detail2 Detail3 2A Line2 Detail 3A Line3 Detail Detail
And want to extract these, knowing that each line starts with #A Line# (where # is a number):
1A Line1 Detail1 Detail2 Detail3
2A Line2 Detail
3A Line3 Detail Detail
You might try this, which fails because of greediness...
[0-9]+A Line[0-9]+(?: \w+)+
Or this, which when made lazy no more works...
[0-9]+A Line[0-9]+(?: \w+)+?
But with a positive lookahead, you get this:
[0-9]+A Line[0-9]+(?: \w+)+?(?= [0-9]+A Line[0-9]+|$)
And appropriately extracts what's needed.
Another possible situation is one where you have something like this:
#ff00fffirstword#445533secondword##008877thi#rdword#
Which you want to convert to three pairs of variables (first of the pair being a # and some hex values (6) and whatever characters after them):
#ff00ff and firstword
#445533 and secondword#
#008877 and thi#rdword#
If there were no hashes inside the 'words', it would have been enough to use (#[0-9a-f]{6})([^#]+), but unfortunately, that's not the case and you have to resort to .*? instead of [^#]+, which doesn't quite yet solve the issue of stray hashes. Positive lookaheads however make this possible:
(#[0-9a-f]{6})(.+?)(?=#[0-9a-f]{6}|$)
Validation & Formatting
Not recommended, but you can use positive lookaheads for quick validations. The following regex for instance allow the entry of a string containing at least 1 digit and 1 lowercase letter.
^(?=[^0-9]*[0-9])(?=[^a-z]*[a-z])
This can be useful when you're checking for character length but have patterns of varying length in the a string, for example, a 4 character long string with valid formats where # indicates a digit and the hyphen/dash/minus - must be in the middle:
##-#
#-##
A regex like this does the trick:
^(?=.{4}$)\d+-\d+
Where otherwise, you'd do ^(?:[0-9]{2}-[0-9]|[0-9]-[0-9]{2})$ and imagine now that the max length was 15; the number of alterations you'd need.
If you want a quick and dirty way to rearrange some dates in the 'messed up' format mmm-yyyy and yyyy-mm to a more uniform format mmm-yyyy, you can use this:
(?=.*(\b\w{3}\b))(?=.*(\b\d{4}\b)).*
Input:
Oct-2013
2013-Oct
Output:
Oct-2013
Oct-2013
An alternative might be to use a regex (normal match) and process separately all the non-conforming formats separately.
Something else I came across on SO was the indian currency format, which was ##,##,###.### (3 digits to the left of the decimal and all other digits groupped in pair). If you have an input of 122123123456.764244, you expect 1,22,12,31,23,456.764244 and if you want to use a regex, this one does this:
\G\d{1,2}\K\B(?=(?:\d{2})*\d{3}(?!\d))
(The (?:\G|^) in the link is only used because \G matches only at the start of the string and after a match) and I don't think this could work without the positive lookahead, since it looks forward without moving the point of replacement.)
Trimming
Suppose you have:
this is a sentence
And want to trim all the spaces with a single regex. You might be tempted to do a general replace on spaces:
\s+
But this yields thisisasentence. Well, maybe replace with a single space? It now yields " this is a sentence " (double quotes used because backticks eats spaces). Something you can however do is this:
^\s*|\s$|\s+(?=\s)
Which makes sure to leave one space behind so that you can replace with nothing and get "this is a sentence".
Splitting
Well, somewhere else where positive lookarounds might be useful is where, say you have a string ABC12DE3456FGHI789 and want to get the letters+digits apart, that is you want to get ABC12, DE3456 and FGHI789. You can easily do use the regex:
(?<=[0-9])(?=[A-Z])
While if you use ([A-Z]+[0-9]+) (i.e. the captured groups are put back in the resulting list/array/etc, you will be getting empty elements as well.
Note that this could be done with a match as well, with [A-Z]+[0-9]+
If I had to mention negative lookarounds, this post would have been even longer :)
Keep in mind that a positive/negative lookaround is the same for a regex engine. The goal of lookarounds is to perform a check somewhere in your "regular expression".
One of the main interest is to capture something without using capturing parenthesis (capturing the whole pattern), example:
string: aaabbbccc
regex: (?<=aaa)bbb(?=ccc)
(you obtain the result with the whole pattern)
instead of: aaa(bbb)ccc
(you obtain the result with the capturing group.)