I have a very basic regular expression that I just can't figure out why it's not working so the question is two parts. Why does my current version not work and what is the correct expression.
Rules are pretty simple:
Must have minimum 3 characters.
If a % character is the first character must be a minimum of 4 characters.
So the following cases should work out as follows:
AB - fail
ABC - pass
ABCDEFG - pass
% - fail
%AB - fail
%ABC - pass
%ABCDEFG - pass
%%AB - pass
The expression I am using is:
^%?\S{3}
Which to me means:
^ - Start of string
%? - Greedy check for 0 or 1 % character
\S{3} - 3 other characters that are not white space
The problem is, the %? for some reason is not doing a greedy check. It's not eating the % character if it exists so the '%AB' case is passing which I think should be failing. Why is the %? not eating the % character?
Someone please show me the light :)
Edit: The answer I used was Dav below: ^(%\S{3}|[^%\s]\S{2})
Although it was a 2 part answer and Alan's really made me understand why. I didn't use his version of ^(?>%?)\S{3} because it worked but not in the javascript implementation. Both great answers and a lot of help.
The word for the behavior you described isn't greedy, it's possessive. Normal, greedy quantifiers match as much as they can originally, but back off if necessary to allow the whole regex to match (I like to think of them as greedy but accommodating). That's what's happening to you: the %? originally matches the leading percent sign, but if there aren't enough characters left for an overall match, it gives up the percent sign and lets \S{3} match it instead.
Some regex flavors (including Java and PHP) support possessive quantifiers, which never back off, even if that causes the overall match to fail. .NET doesn't have those, but it has the next best thing: atomic groups. Whatever you put inside an atomic group acts like a separate regex--it either matches at the position where it's applied or it doesn't, but it never goes back and tries to match more or less than it originally did just because the rest of the regex is failing (that is, the regex engine never backtracks into the atomic group). Here's how you would use it for your problem:
^(?>%?)\S{3}
If the string starts with a percent sign, the (?>%?) matches it, and if there aren't enough characters left for \S{3} to match, the regex fails.
Note that atomic groups (or possessive quantifiers) are not necessary to solve this problem, as #Dav demonstrated. But they're very powerful tools which can easily make the difference between impossible and possible, or too damn slow and slick as can be.
Regex will always try to match the whole pattern if it can - "greedy" doesn't mean "will always grab the character if it exists", but instead means "will always grab the character if it exists and a match can be made with it grabbed".
Instead, what you probably want is something like this:
^(%\S{3}|[^%\s]\S{2})
Which will match either a % followed by 3 characters, or a non-%, non-whitespace followed by 2 more.
I always love to look at RE questions to see how much time people spend on them to "Save time"
str.len() >= str[0]=='&' ? 4 : 3
Although in real life I'd be more explicit, I just wrote it that way because for some reason some people consider code brevity an advantage (I'd call it an anti-advantage, but that's not a popular opinion right now)
Try the regex modified a little based on Dav's original one:
^(%\S{3,}|[^%\s]\S{2,})
with the regex option "^ and $ match at line breaks" on.
Related
I'm trying to create a regex for "#h|H\s#min|MIN" format case insensitive For example 1H 2MIn, 2h 03min should match. 1H 60MIN, 02h 100min should not match. Thanks to Jesse point out. There is no limit on how many digits in hours. 60min suppose to be an hour. So anything above 59mins should not match.
Currently, I got:
/^[0-9]H|h\s([0-5]?\d)(MIN)|(min)$/
It's not worked for cases like 02h 100min.
So which part I did do wrong?
I thought ([0-5]?\d) suppose only match two digits.
Thank you for any help!
Edit:
I think I figured this out.
/^\d+h\s[0-5]?\dmin$/i
worked in this case.
Thanks again
The Problem
The main reason your expression isn't working as expected is because of how you're using the alternative (|) operator. This operator tells the engine that it can either match the pattern on the left side, or the pattern on the right side, but not both. The problem here is that these options aren't confined to a group, so your options aren't what you think they are.
Your entire expression is actually split into three alternative expressions. The engine is being told to match:
^[0-9]H OR
h\s([0-5]?\d)(MIN) OR
(min)$
This explains the strange behavior. This can be fixed by confining what you actually want to check for to a group. In this case, since you're wanting to match H or h and MIN or min, we can create the groups (H|h) and (MIN|min). Pretty simple. After these changes your expression would look like this:
^[0-9](H|h)\s([0-5]?\d)(MIN|min)$
Regex101
However this expression has another problem. It only matches one digit in the hour section. This can be fixed by adding a quantifier to it. In this case we can use +. This tells the engine to match the previous token between 1 and unlimited times. We can apply this to [0-9] to get [0-9]+, which will match any number between 0 and 9 between 1 and unlimited times. After this change your expression would look like this:
^[0-9]+(H|h)\s([0-5]?\d)(MIN|min)$
Regex101
Now you have a working expression, but it can be simplified quite a bit, so let's talk about that next.
Improvements
First, [0-9] can be replaced with \d. These do exactly the same thing. They both match any digit between 0 and 9.
^\d+(H|h)\s([0-5]?\d)(MIN|min)$
Regex101
Second, since this problem was caused by your need to match both upper and lower case, let's eliminate this problem entirely by using the case insensitivity (i) flag. The easiest way to apply this is by using (?i) at the beginning of your expression. This works across every language that supports Regex. With this flag, we no longer have to worry about matching upper and lower case letters. We can replace (H|h) with just h, and (MIN|min) with just min.
(?i)^\d+h\s([0-5]?\d)min$
Regex101
This expression should do the job. You can save 3 more characters by replacing \s with (space) and by removing the parentheses around [0-5]?\d, but those are insignificant changes so I'll leave that up to you.
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$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
My string being of the form:
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
I only want to match against the last segment of whitespace before the last period(.)
So far I am able to capture whitespace but not the very last occurrence using:
\s+(?=\.\w)
How can I make it less greedy?
In a general case, you can match the last occurrence of any pattern using the following scheme:
pattern(?![\s\S]*pattern)
(?s)pattern(?!.*pattern)
pattern(?!(?s:.*)pattern)
where [\s\S]* matches any zero or more chars as many as possible. (?s) and (?s:.) can be used with regex engines that support these constructs so as to use . to match any chars.
In this case, rather than \s+(?![\s\S]*\s), you may use
\s+(?!\S*\s)
See the regex demo. Note the \s and \S are inverse classes, thus, it makes no sense using [\s\S]* here, \S* is enough.
Details:
\s+ - one or more whitespace chars
(?!\S*\s) - that are not immediately followed with any 0 or more non-whitespace chars and then a whitespace.
You can try like so:
(\s+)(?=\.[^.]+$)
(?=\.[^.]+$) 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+)\.[^\.]*$
I have this regex
<#([^\s]+).*?>\s?<a href=""(.*?)"".*?>(.*?)</a>(\s?\((Pending|Prepared)\))?
And i really need it in a vba version for words .find method (don't need the matching-groups), here is what i have so far
\<\#*\>*\<a href=*\>*\<\/a\>
But i cant get the last part to work, here I'm talking about
(\s?\((Pending|Prepared)\))?
I really hope someone can help me, as regex in this case is not an option (Although i know i can use regex in VBA!)
Cheers
I don't see an OR | in the documentation (Wildcard character reference) or the examples (Putting regular expressions to work in Word), so instead I suggest splitting it into two separate searches. The Word MVPs site has a good reference on the Word Regex as well if you want more information.
[^\s] can be written in the Word style regex as [! ] (note the space), + becomes #. It appears that neither the {n,} nor {n,m} syntax of VBA support an n value of 0, making ? and * hard to implement in Word. One option that the MS guys seem to use is *, which in Word is "Any string of characters". By my testing, * is lazy, meaning the pattern \<#*\> run against the string <#sometag> asdfsadfasdf > will only match <#sometag>. In addition, it can match 0 characters, for example \<\#*\> will match <#>.
So assuming that the first part is working as you expect, you could try the following two regex:
\<\#*\>*\<a href=*\>*\<\/a\>*\(Pending\)
and
\<\#*\>*\<a href=*\>*\<\/a\>*\(Prepared\)
The trouble here is that the * will match up until it hits the P of Pending or Prepared, so there could be other text in between, but it's the only way I can see of matching an optional space. If you can guaruntee that the space will or will not be there, that would go a long way towards making the regex safer.
Give that a try and see if it works for you!
One of my homework questions asked to develop a regex for all strings over x,y,z that did not contain xxx
After doing some reading I found out about negative lookahead and made this which works great:
(x(?!xx)|y|z)*
Still, in the spirit of completeness, is there anyway to write this without negative lookahead?
Reading I have done makes me think it can be done with some combination of carets (^), but I cannot get the right combination so I am not sure.
Taking it a step further, is it possible to exclude a string like xxx using only the or (|) operator, but still check the strings in a recursive fashion?
EDIT 9/6/2010:
Think I answered my own question. I messed with this some more, trying make this regex with only or (|) statements and I am pretty sure I figured it out... and it isn't nearly as messy as I thought it would be. If someone else has time to verify this with a human eye I would appreciate it.
(xxy|xxz|xy|xz|y|z)*(xxy|xxz|xx|xy|xz|x|y|z)
Try this:
^(x{0,2}(y|z|$))*$
The basic idea is this: for match at most 2 X's, followed by another letter or the end of the string.
When you reach a point where you have 3 X's, the regex has no rule that allows it to keep matching, and it fails.
Working example: http://rubular.com/r/ePH0fHlZxL
A less compact way to write the same is (with free spaces, usually the /x flag):
^(
y| # y is ok
z| # so is z
x(y|z|$)| # a single x, not followed by x
xx(y|z|$) # 2 x's, not followed by x
)*$
Based on the latest edit, here's an ever flatter version of the pattern: I'm not entirely sure I understand your fascination with the pipe, but you can eliminate some more options - by allowing an empty match on the second group you don't need to repeat permutations from the first group. That regex also allows ε, which I think is included in your language.
^(xxy|xxz|xy|xz|y|z)*(xx|x|)$
Basically you have the right answer already - well done you. :)
Carat (^) in a set [^abc] will only match where it does not find a character in that set so it's application for matching orders of characters (i.e. strings) is limited and weak.
Regex has numeric quantifiers {n} and {a,b} which allow you to match a defined number of repititions of a pattern, which would work for this specific pattern (because it's 'x' repeated) but it's not particularily expressive of the problem you're trying to solve (even for regex!) and is a bit brittle (it wouldn't be appropriate for negative match 'xyx' for example.
An or pattern again would be verbose and rather unexpressive but it could be done as the fragment:
(x|xx)[^x] // x OR xx followed by NOT x
Obviously you can do this with an iterative algorithm but that's highly inefficient compared to a regex.
Well done for thinking beyond the solution though.
I know you don't want to use lookahead, but here's another way to solve this:
^(?:(?!xxx)[xyz])*$
will match any line of characters x, y or z as long as it doesn't contain the string xxx.