I'm trying to write a regex that would match groups of exactly three characters, that reoccur within the text at least one time.
What I came up with is this simple regex:(.{3}).*\g1, using the \g (global) and \s (dot also matches newline) flags. However, it is clearly faulty, as it only finds a part of the groups I'm hoping to capture. Any idea how can I improve it? Here is the link to an example input https://regex101.com/r/Cuiva1/2
Edit: Here's the full list of groups I was hoping to capture as requested in the comment:GLT,VIW,IWK,KTL,GLT,LTK,LIS,KTX,TXK,XDL,KTL
If your input is always multiple triplets of uppercase characters and you're only looking for ones that repeat, then you need something more complex to avoid backtracking into a previous triplet:
/(?>[^A-Z]*+([A-Z]{3}))(?=(?:[^A-Z]*+[A-Z]{3})*?\1)|(?>[^A-Z]*+[A-Z]{3})/g
The matches from index 1 will hold what you want. If your strings are not that well formatted (i.e. may contain any length string in between repeating patterns, then you can use a simpler pattern but you'll get totally inconsistent results and miss some matches.
I re-read your desired output, you're not going to achieve this with regex. VIW and IWK are overlapping, which won't work in a single preg_match_all(). Just use string functions.
Related
I am trying to regex the following string:
https://www.amazon.com/Tapps-Top-Apps-and-Games/dp/B00VU2BZRO/ref=sr_1_3?ie=UTF8&qid=1527813329&sr=8-3&keywords=poop
I want only B00VU2BZRO.
This substring is always going to be a 10 characters, alphanumeric, preceded by dp/.
So far I have the following regex:
[d][p][\/][0-9B][0-9A-Z]{9}
This matches dp/B00VU2BZRO
I want to match only B00VU2BZRO with no dp/
How do I regex this?
Here is one regex option which would produce an exact match of what you want:
(?<=dp\/)(.*)(?=\/)
Demo
Note that this solution makes no assumptions about the length of the path fragment occurring after dp/. If you want to match a certain number of characters, replace (.*) with (.{10}), for example.
Depending on your language/method of application, you have a couple of options.
Positive look behind. This will make your regex more complicated, but will make it match what you want exactly:
(<=dp/)[0-9A-Z]{10}
The construct (<=...) is called a positive look behind. It will not consume any of the string, but will only allow the match to happen if the pattern between the parens is matched.
Capture group. This will make the regex itself slightly simpler, but will add a step to the extraction process:
dp/([0-9A-Z]{10})
Anything between plain parens is a capture group. The entire pattern will be matched, including dp/, but most languages will give you a way of extracting the portion you are interested in.
Depending on your language, you may need to escape the forward slash (/).
As an aside, you never need to create a character class for single characters: [d][p][\/] can equally well be written as just dp\/.
I need some help with building up my regex.
What I am trying to do is match a specific part of text with unpredictable parts in between the fixed words. An example is the sentence one gets when replying to an email:
On date at time person name has written:
The cursive parts are variable, might contains spaces or a new line might start from this point.
To get this, I built up my regex as such: On[\s\S]+?at[\s\S]+?person[\s\S]+?has written:
Basically, the [\s\S]+? is supposed to fill in any letter, number, space or break/new line as I am unable to predict what could be between the fixed words tha I am sure will always be there.
Now comes the hard part, when I would add the word "On" somewhere in the text above the sentence that I want to match, the regex now matches a much bigger text than I want. This is due to the use of [\s\S]+.
How am I able to make my regex match as less characters as possible? Using "?" before the "+" to make it lazy does not help.
Example is here with words "From - This - Point - Everything:". Cases are ignored.
Correct: https://regexr.com/3jdek.
Wrong because of added "From": https://regexr.com/3jdfc
The regex is to be used in VB.NET
A more real life, with html tags, can be found here. Here, I avoided using [\s\S]+? or (.+)?(\r)?(\n)?(.+?)
Correct: https://regexr.com/3jdd1
Wrong: https://regexr.com/3jdfu after adding certain parts of the regex in the text above. Although, in html, barely possible to occur as the user would never write the matching tag himself, I do want to make sure my regex is correctjust in case
These things are certain: I know with what the part of text starts, no matter where in respect to the entire text, I know with what the part of text ends, and there are specific fixed words that might make the regex more reliable, but they can be ommitted. Any text below the searched part is also allowed to be matched, but no text above may be matched at all
Another example where it goes wrong: https://regexr.com/3jdli. Basically, I have less to go with in this text, so the regex has less tokens to work with. Adding just the first < already makes the regex take too much.
From my own experience, most problems are avoided when making sure I do not use any [\s\S]+? before I did a (\r)?(\n)? first
[\s\S] matches all character because of union of two complementary sets, it is like . with special option /s (dot matches newlines). and regex are greedy by default so the largest match will be returned.
Following correct link, the token just after the shortest match must be geschreven, so another way to write without using lazy expansion, which is more flexible is to prepend the repeated chracter set by a negative lookahead inside loop,
so
<blockquote type="cite" [^>]+?>[^O]+?Op[^h]+?heeft(.+?(?=geschreven))geschreven:
becomes
<blockquote type="cite" [^>]+?>[^O]+?Op[^h]+?heeft((?:(?!geschreven).)+)geschreven:
(?: ) is for non capturing the group which just encapsulates the negative lookahead and the . (which can be replaced by [\s\S])
(?! ) inside is the negative lookahead which ensures current position before next character is not the beginning of end token.
Following comments it can be explicitly mentioned what should not appear in repeating sequence :
From(?:(?!this)[\s\S])+this(?:(?!point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
or
From(?:(?!From|this)[\s\S])+this(?:(?!point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
or
From(?:(?!From|this)[\s\S])+this(?:(?!this|point)[\s\S])+point(?:(?!everything)[\s\S])+everything:
to understand what the technic (?:(?!tokens)[\s\S])+ does.
in the first this can't appear between From and this
in the second From or this can't appear between From and this
in the third this or point can't appear between this and point
etc.
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 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!
Recently I'm writing/testing regexps on https://regex101.com/.
My question is: Is it possible to do a positive look-ahead AND a replacement in the same "replacement"? Or just limited kind of replacement is possible.
Input is several lines with phone numbers. Let's say the correct phone number where the number of "numbers" are 11. No matter how the numbers are divided/group together with - / characters, no matter if starts with + 00 or it is omitted.
Some example lines:
+48301234567
+48/30/1234567
+48-30-12-345-67
+483011223344556677
0048301234567
+(48)30/1234567
Positive look-ahead able to check if from the beginning until the end of line there are only 11 digits, regardless how many other, above specified character separating them. This works perfectly.
Where the positive look-ahead check is fine, I would like to delete every character but numbers. The replacement works fine until I'm not involving look-ahead.
Checking the regexp itself working perfectly ("gm" modes):
^(?:\+|00)?(?:[\-\/\(\)]?\d){11}$
Checking the replace part works perfectly (replace to nothing):
[^\d\n]
Put this into look-ahead, after the deletion of non new-line and non-digit characters from the matching lines:
(?=^(?:\+|00)?(?:[\-\/\(\)]?\d){11}$)[^\d\n]
Even I put the ^ $ into look-ahead, seems the replacement working only from beginning of the lines until the very first digit.
I know in real life the replacement and the check should/would go separate ways, however I'm curious if I could mix look-ahead/look-behind with string operations like replace, delete, take the string apart and put together as I like.
UPDATE: This is what would do the trick, however I feel this one "ugly" a bit. Is there any prettier solution?
https://regex101.com/r/yT5dA4/2
Or the version which I asked originally, where only digits remains: regex101.com/r/yT5dA4/3
You cannot replace/delete text with regex. Regex is just a tool for matching certain strings and then taking certain action depending on the matching text, eg. perform a substitution, retrieve the second capture group.
However it is possible to perform certain decisions within a regex engine, by using conditionals. The common syntax for this, with a lookahead assertion, is (?(?=regex)then|else).
With conditionals you can change the behaviour depending on how the text matches the regex. For your example you could do something like:
^(\+)?(?(1)\(|\d)
If the phone number starts with a plus it must be followed by a bracket, else it should start with a digit. Although in your situation, this is not very useful.
If you want to read up more on conditionals in regex you can do so here.