I am having trouble understanding negative regex lookahead / lookbehind. I got the impression from reading tutorials that when you set a criteria to look for, the criteria doesn't form part of the search match.
That seems to hold for positive lookahead examples I tried, but when I tried these negative ones, it matches the entire test string. 1, it shouldn't have matched anything, and 2 even if it did, it wasn't supposed to include the lookahead criteria??
(?<!^And).*\.txt$
with input
And.txt
See: https://regex101.com/r/vW0aXS/1
and
^A.*(?!\.txt$)
with input:
A.txt
See: https://regex101.com/r/70yeED/1
PS: if you're going to ask me which language. I don't know. we've been told to use regex without any specific reference to any specific languages. I tried clicking various options on regex101.com and they all came up the same.
Lookarounds only try to match at their current position.
You are using a lookbehind at the beginning of the string (?<!^And).*\.txt$, and a lookahead at the end of the string ^A.*(?!\.txt$), which won't work. (.* will always consume the whole string as it's first match)
To disallow "And", for example, you can put the lookahead at the beginning of the string with a greedy quantifier .* inside it, so that it scans the whole string:
(?!.*And).*\.txt$
https://regex101.com/r/1vF50O/1
Your understanding is correct and the issue is not with the lookbehind/lookahead. The issue is with .* which matches the entire string in both cases. The period . matches any character and then you follow it with * which makes it match the entire string of any length. Remove it and both you regexes will work:
(?<!^And)\.txt$
^A(?!\.txt$)
Related
I've searched through multiple answers on SO now, but most of them consider the beginning of the line as the whole string being looked upon, which doesn't serve my case, I think (at least all the answers I tried didn't work).
So, I want to match all codes within a text that are 7-digit long, start with 1 or 2, and are not prefixed by "TC-" and its lowercase variants.
Came up with the /(!?TC-){0}(1|2)\d{6}/g expression, but it doesn't work for not matching the codes that start with "TC-", and I don't know how can I prevent from selecting those. Is there a way to do that?
I've created an example pattern on Regexr: regexr.com/6p70c.
You can assert not TC- to the left using negative lookbehind (?<! and omit the {0} quantifier as that makes it optional:
(?<!\bTC-)\b[12]\d{6}\b
Regex demo
I am having issues excluding parts of a string in a VSCode Snippet. Essentially, what I want is a specific piece of a path but I am unable to get the regex to exclude what I need excluded.
I have recently asked a question about something similar which you can find here: Is there a way to trim a TM_FILENAME beyond using TM_FILENAME_BASE?
As you can see, I am getting mainly tripped up by how the snippets work within vscode and not so much the regular expressions themselves
${TM_FILEPATH/(?<=area)(.+)(?=state)/${1:/pascalcase}/}
Given a file path that looks like abc/123/area/my-folder/state/...
Expected:
/MyFolder/
Actual:
abc/123/areaMyFolderstate/...
You need to match the whole string to achieve that:
"${TM_FILEPATH/.*area(\\/.*?\\/)state.*/${1:/pascalcase}/}"
See the regex demo
Details
.* - any 0+ chars other than line break chars, as many as possible
area - a word
-(\\/.*?\\/) - Group 1: /, any 0+ chars other than line break chars, as few as possible, and a /
-state.* - state substring and the rest of the line.
NOTE: If there must be no other subparts between area and state, replace .*? with [^\\/]* or even [^\\/]+.
The expected output seems to be different with part of a string in the input. If that'd be desired the expression might be pretty complicated, such as:
(?:[\s\S].*?)(?<=area\/)([^-])([^-]*)(-)([^\/])([^\/]*).*
and a replacement of something similar to /\U$1\E$2$3\U$4\E$5/, if available.
Demo 1
If there would be other operations, now I'm guessing maybe the pascalcase would do something, this simple expression might simply work here:
.*area(\\/.*?\\/).*
and the desired data is in this capturing group $1:
(\\/.*?\\/)
Demo 2
Building on my answer you linked to in your question, remember that lookarounds are "zero-length assertions" and "do not consume characters in the string". See lookarounds are zero-length assertions:
Lookahead and lookbehind, collectively called "lookaround", are zero-length assertions just like the start and end of line, and start and end of word anchors explained earlier in this tutorial. The difference is that lookaround actually matches characters, but then gives up the match, returning only the result: match or no match. That is why they are called "assertions". They do not consume characters in the string, but only assert whether a match is possible or not.
So in your snippet transform: /(?<=area)(.+)(?=state)/ the lookaround portions are not actually consumed and so are simply passed through. Vscode treats them, as it should, as not actually being within the "part to be transformed" segment at all.
That is why lookarounds are not excluded from your transform.
I was browsing through the regex tagged questions on SO when i came accross this problem,
A regex for a url was needed, the url begins with domain.com/advertorials/
The regex should match the following scenarios,
domain.com/advertorials
domain.com/advertorials?test=true
domain.com/advertorials/
domain.com/advertorials/?test=true
but not this,
domain.com/advertorials/version1?test=true
I came up with this regex advertorials\/?(?:(?!version)(.*))
This should work, but it doesnt for the last case. Looking at the debugger in regex101.com,
i see that after matching 's/' it matches 'version' word character by character and ultimately matches but since this is negative lookahead the condition fails. And this is the part i dont understand after failing it backtracks to before the '/' in 's/' and not after 's/'.
Is this how its supposed to work?? Can anyone help me understand?
(here's the demo link: https://regex101.com/r/ww3HR8/1).
Thanks,
Note: People already gave their solutions on that problem i just want to know why my regex fails.
The backtracking mechanism is in charge of this phenomenon, as you have already pointed out.
The ? quantifier, matching 1 or 0 repetitions of the quantified subpattern lets the regex engine match the string in two ways: either matching the quantified subpattern, or go on matching the string with subsequent subpattern.
So, advertorials/?(?!version)(.*) (I removed the redundant (?:...) non-capturing group), when applied to domain.com/advertorials/version1?test=true, matches advertorials, then matches /, and then the negative lookahead checks if, immediately to the right of the current position, there is version substring. Since there is version after /, the regex engine goes back and sees that /? pattern can match an empty string. So, the lookahead check is re-applied striaght after advertorials. There is no version after advertorials, and the match is returned.
The usual solution is using possessive quantifiers or atomic groups, but there are other approaches, too.
E.g.
advertorials\/?+(?!version)(.*)
^^
See the regex demo. Here, \/?+ matches 1 or 0 / chars, but once it matches, the egine cannot go back and re-match a part of a string with this pattern.
Or, you may include the /? in the lookahead and place it before /? pattern:
advertorials(?!\/?version)\/?(.*)
See another regex demo.
If you plan to disallow version anywhere after advertorials use
advertorials(?!.*version)\/?(.*)
See yet another demo.
Making the slash optional means there is a way to match without violating the constraint. If there is a way to match, the regex engine will find it, always.
Make the slash non-optional when it's followed by anything at all.
advertorials(?:/(?!version).*)?$
Incidentally, regex itself doesn't require the slash to be backslash-escaped (though some host languages use slashes as regex delimiters, so maybe you need to put it back). I also removed some redundant parentheses.
The reason:
This highlighted part is optional
advertorials\/?(?:(?!version)(.*))
Therefore it can also be advertorials(?:(?!version)(.*))
which matches advertorials/version
Essentially, (?!version)(.*) matches /version
Btw, this is normal backtracking by 1 character.
If you have already fixed it, then we're done !
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+)\.[^\.]*$
Let's say I have a regular expression that works correctly to find all of the URLs in a text file:
(http://)([a-zA-Z0-9\/\.])*
If what I want is not the URLs but the inverse - all other text except the URLs - is there an easy modification to make to get this?
You could simply search and replace everything that matches the regular expression with an empty string, e.g. in Perl s/(http:\/\/)([a-zA-Z0-9\/\.])*//g
This would give you everything in the original text, except those substrings that match the regular expression.
If for some reason you need a regex-only solution, try this:
((?<=http://[a-zA-Z0-9\/\.#?/%]+(?=[^a-zA-Z0-9\/\.#?/%]))|\A(?!http://[a-zA-Z0-9\/\.#?/%])).+?((?=http://[a-zA-Z0-9\/\.#?/%])|\Z)
I expanded the set of of URL characters a little ([a-zA-Z0-9\/\.#?/%]) to include a few important ones, but this is by no means meant to be exact or exhaustive.
The regex is a bit of a monster, so I'll try to break it down:
(?<=http://[a-zA-Z0-9\/\.#?/%]+(?=[^a-zA-Z0-9\/\.#?/%])
The first potion matches the end of a URL. http://[a-zA-Z0-9\/\.#?/%]+ matches the URL itself, while (?=[^a-zA-Z0-9\/\.#?/%]) asserts that the URL must be followed by a non-URL character so that we are sure we are at the end. A lookahead is used so that the non-URL character is sought but not captured. The whole thing is wrapped in a lookbehind (?<=...) to look for it as the boundary of the match, again without capturing that portion.
We also want to match a non-URL at the beginning of the file. \A(?!http://[a-zA-Z0-9\/\.#?/%]) matches the beginning of the file (\A), followed by a negative lookahead to make sure there's not a URL lurking at the start of the file. (This URL check is simpler than the first one because we only need the beginning of the URL, not the whole thing.)
Both of those checks are put in parenthesis and OR'd together with the | character. After that, .+? matches the string we are trying to capture.
Then we come to ((?=http://[a-zA-Z0-9\/\.#?/%])|\Z). Here, we check for the beginning of a URL, once again with (?=http://[a-zA-Z0-9\/\.#?/%]). The end of the file is also a pretty good sign that we've reached the end of our match, so we should look for that, too, using \Z. Similarly to a first big group, we wrap it in parenthesis and OR the two possibilities together.
The | symbol requires the parenthesis because its precedence is very low, so you have to explicitly state the boundaries of the OR.
This regex relies heavily on zero-width assertions (the \A and \Z anchors, and the lookaround groups). You should always understand a regex before you use it for anything serious or permanent (otherwise you might catch a case of perl), so you might want to check out Start of String and End of String Anchors and Lookahead and Lookbehind Zero-Width Assertions.
Corrections welcome, of course!
If I understand the question correctly, you can use search/replace...just wildcard around your expression and then substitute the first and last parts.
s/^(.*)(your regex here)(.*)$/$1$3/
im not sure if this will work exactly as you intend but it might help:
Whatever you place in the brackets [] will be matched against. If you put ^ within the bracket, i.e [^a-zA-Z0-9/.] it will match everything except what is in the brackets.
http://www.regular-expressions.info/