Related
https://regex101.com/r/sB9wW6/1
(?:(?<=\s)|^)#(\S+) <-- the problem in positive lookbehind
Working like this on prod: (?:\s|^)#(\S+), but I need a correct start index (without space).
Here is in JS:
var regex = new RegExp(/(?:(?<=\s)|^)#(\S+)/g);
Error parsing regular expression: Invalid regular expression:
/(?:(?<=\s)|^)#(\S+)/
What am I doing wrong?
UPDATE
Ok, no lookbehind in JS :(
But anyways, I need a regex to get the proper start and end index of my match. Without leading space.
Make sure you always select the right regex engine at regex101.com. See an issue that occurred due to using a JS-only compatible regex with [^] construct in Python.
JS regex - at the time of answering this question - did not support lookbehinds. Now, it becomes more and more adopted after its introduction in ECMAScript 2018. You do not really need it here since you can use capturing groups:
var re = /(?:\s|^)#(\S+)/g;
var str = 's #vln1\n#vln2\n';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
The (?:\s|^)#(\S+) matches a whitespace or the start of string with (?:\s|^), then matches #, and then matches and captures into Group 1 one or more non-whitespace chars with (\S+).
To get the start/end indices, use
var re = /(\s|^)#\S+/g;
var str = 's #vln1\n#vln2\n';
var pos = [];
while ((m = re.exec(str)) !== null) {
pos.push([m.index+m[1].length, m.index+m[0].length]);
}
console.log(pos);
BONUS
My regex works at regex101.com, but not in...
First of all, have you checked the Code Generator link in the Tools pane on the left?
All languages - "Literal string" vs. "String literal" alert - Make sure you test against the same text used in code, literal string, at the regex tester. A common scenario is copy/pasting a string literal value directly into the test string field, with all string escape sequences like \n (line feed char), \r (carriage return), \t (tab char). See Regex_search c++, for example. Mind that they must be replaced with their literal counterparts. So, if you have in Python text = "Text\n\n abc", you must use Text, two line breaks, abc in the regex tester text field. Text.*?abc will never match it although you might think it "works". Yes, . does not always match line break chars, see How do I match any character across multiple lines in a regular expression?
All languages - Backslash alert - Make sure you correctly use a backslash in your string literal, in most languages, in regular string literals, use double backslash, i.e. \d used at regex101.com must written as \\d. In raw string literals, use a single backslash, same as at regex101. Escaping word boundary is very important, since, in many languages (C#, Python, Java, JavaScript, Ruby, etc.), "\b" is used to define a BACKSPACE char, i.e. it is a valid string escape sequence. PHP does not support \b string escape sequence, so "/\b/" = '/\b/' there.
All languages - Default flags - Global and Multiline - Note that by default m and g flags are enabled at regex101.com. So, if you use ^ and $, they will match at the start and end of lines correspondingly. If you need the same behavior in your code check how multiline mode is implemented and either use a specific flag, or - if supported - use an inline (?m) embedded (inline) modifier. The g flag enables multiple occurrence matching, it is often implemented using specific functions/methods. Check your language reference to find the appropriate one.
line-breaks - Line endings at regex101.com are LF only, you can't test strings with CRLF endings, see regex101.com VS myserver - different results. Solutions can be different for each regex library: either use \R (PCRE, Java, Ruby) or some kind of \v (Boost, PCRE), \r?\n, (?:\r\n?|\n)/(?>\r\n?|\n) (good for .NET) or [\r\n]+ in other libraries (see answers for C#, PHP). Another issue related to the fact that you test your regex against a multiline string (not a list of standalone strings/lines) is that your patterns may consume the end of line, \n, char with negated character classes, see an issue like that. \D matched the end of line char, and in order to avoid it, [^\d\n] could be used, or other alternatives.
php - You are dealing with Unicode strings, or want shorthand character classes to match Unicode characters, too (e.g. \w+ to match Стрибижев or Stribiżew, or \s+ to match hard spaces), then you need to use u modifier, see preg_match() returns 0 although regex testers work - To match all occurrences, use preg_match_all, not preg_match with /...pattern.../g, see PHP preg_match to find multiple occurrences and "Unknown modifier 'g' in..." when using preg_match in PHP?- Your regex with inline backreference like \1 refuses to work? Are you using a double quoted string literal? Use a single-quoted one, see Backreference does not work in PHP
phplaravel - Mind you need the regex delimiters around the pattern, see https://stackoverflow.com/questions/22430529
python - Note that re.search, re.match, re.fullmatch, re.findall and re.finditer accept the regex as the first argument, and the input string as the second argument. Not re.findall("test 200 300", r"\d+"), but re.findall(r"\d+", "test 200 300"). If you test at regex101.com, please check the "Code Generator" page. - You used re.match that only searches for a match at the start of the string, use re.search: Regex works fine on Pythex, but not in Python - If the regex contains capturing group(s), re.findall returns a list of captures/capture tuples. Either use non-capturing groups, or re.finditer, or remove redundant capturing groups, see re.findall behaves weird - If you used ^ in the pattern to denote start of a line, not start of the whole string, or used $ to denote the end of a line and not a string, pass re.M or re.MULTILINE flag to re method, see Using ^ to match beginning of line in Python regex
- If you try to match some text across multiple lines, and use re.DOTALL or re.S, or [\s\S]* / [\s\S]*?, and still nothing works, check if you read the file line by line, say, with for line in file:. You must pass the whole file contents as the input to the regex method, see Getting Everything Between Two Characters Across New Lines. - Having trouble adding flags to regex and trying something like pattern = r"/abc/gi"? See How to add modifers to regex in python?
c#, .net - .NET regex does not support possessive quantifiers like ++, *+, ??, {1,10}?, see .NET regex matching digits between optional text with possessive quantifer is not working - When you match against a multiline string and use RegexOptions.Multiline option (or inline (?m) modifier) with an $ anchor in the pattern to match entire lines, and get no match in code, you need to add \r? before $, see .Net regex matching $ with the end of the string and not of line, even with multiline enabled - To get multiple matches, use Regex.Matches, not Regex.Match, see RegEx Match multiple times in string - Similar case as above: splitting a string into paragraphs, by a double line break sequence - C# / Regex Pattern works in online testing, but not at runtime - You should remove regex delimiters, i.e. #"/\d+/" must actually look like #"\d+", see Simple and tested online regex containing regex delimiters does not work in C# code - If you unnecessarily used Regex.Escape to escape all characters in a regular expression (like Regex.Escape(#"\d+\.\d+")) you need to remove Regex.Escape, see Regular Expression working in regex tester, but not in c#
dartflutter - Use raw string literal, RegExp(r"\d"), or double backslashes (RegExp("\\d")) - https://stackoverflow.com/questions/59085824
javascript - Double escape backslashes in a RegExp("\\d"): Why do regex constructors need to be double escaped?
- (Negative) lookbehinds unsupported by most browsers: Regex works on browser but not in Node.js - Strings are immutable, assign the .replace result to a var - The .replace() method does change the string in place - Retrieve all matches with str.match(/pat/g) - Regex101 and Js regex search showing different results or, with RegExp#exec, RegEx to extract all matches from string using RegExp.exec- Replace all pattern matches in string: Why does javascript replace only first instance when using replace?
javascriptangular - Double the backslashes if you define a regex with a string literal, or just use a regex literal notation, see https://stackoverflow.com/questions/56097782
java - Word boundary not working? Make sure you use double backslashes, "\\b", see Regex \b word boundary not works - Getting invalid escape sequence exception? Same thing, double backslashes - Java doesn't work with regex \s, says: invalid escape sequence - No match found is bugging you? Run Matcher.find() / Matcher.matches() - Why does my regex work on RegexPlanet and regex101 but not in my code? - .matches() requires a full string match, use .find(): Java Regex pattern that matches in any online tester but doesn't in Eclipse - Access groups using matcher.group(x): Regex not working in Java while working otherwise - Inside a character class, both [ and ] must be escaped - Using square brackets inside character class in Java regex - You should not run matcher.matches() and matcher.find() consecutively, use only if (matcher.matches()) {...} to check if the pattern matches the whole string and then act accordingly, or use if (matcher.find()) to check if there is a single match or while (matcher.find()) to find multiple matches (or Matcher#results()). See Why does my regex work on RegexPlanet and regex101 but not in my code?
scala - Your regex attempts to match several lines, but you read the file line by line (e.g. use for (line <- fSource.getLines))? Read it into a single variable (see matching new line in Scala regex, when reading from file)
kotlin - You have Regex("/^\\d+$/")? Remove the outer slashes, they are regex delimiter chars that are not part of a pattern. See Find one or more word in string using Regex in Kotlin - You expect a partial string match, but .matchEntire requires a full string match? Use .find, see Regex doesn't match in Kotlin
mongodb - Do not enclose /.../ with single/double quotation marks, see mongodb regex doesn't work
c++ - regex_match requires a full string match, use regex_search to find a partial match - Regex not working as expected with C++ regex_match - regex_search finds the first match only. Use sregex_token_iterator or sregex_iterator to get all matches: see What does std::match_results::size return? - When you read a user-defined string using std::string input; std::cin >> input;, note that cin will only get to the first whitespace, to read the whole line properly, use std::getline(std::cin, input); - C++ Regex to match '+' quantifier - "\d" does not work, you need to use "\\d" or R"(\d)" (a raw string literal) - This regex doesn't work in c++ - Make sure the regex is tested against a literal text, not a string literal, see Regex_search c++
go - Double backslashes or use a raw string literal: Regular expression doesn't work in Go - Go regex does not support lookarounds, select the right option (Go) at regex101.com before testing! Regex expression negated set not working golang
groovy - Return all matches: Regex that works on regex101 does not work in Groovy
r - Double escape backslashes in the string literal: "'\w' is an unrecognized escape" in grep - Use perl=TRUE to PCRE engine ((g)sub/(g)regexpr): Why is this regex using lookbehinds invalid in R?
oracle - Greediness of all quantifiers is set by the first quantifier in the regex, see Regex101 vs Oracle Regex (then, you need to make all the quantifiers as greedy as the first one)] - \b does not work? Oracle regex does not support word boundaries at all, use workarounds as shown in Regex matching works on regex tester but not in oracle
firebase - Double escape backslashes, make sure ^ only appears at the start of the pattern and $ is located only at the end (if any), and note you cannot use more than 9 inline backreferences: Firebase Rules Regex Birthday
firebasegoogle-cloud-firestore - In Firestore security rules, the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols, i.e. use allow create: if docId.matches("^\\d+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
Imagine a string that contains special characters like $§%%,., numbers and letters.
I want to receive the letter and number junks of an arbitrary string as an array of strings.
A good solution seems to be the use of regex, but I don't know how to express [numbers and letters]
// example
"abc" = {"abc"};
"ab .c" = {"ab", "c"}
"ab123,cd2, ,,%&$§56" = {"ab123", "cd2", "56"}
// try
string input = "jdahs32455$§&%$§df233§$fd";
string[] output = input.Split(Regex("makejunksfromstring"));
To extract chunks of 1 or more letters/digits you may use
[A-Za-z0-9]+ # ASCII only letters/digits
[\p{L}0-9]+ # Any Unicode letters and ASCII only digits
[\p{L}\p{N}]+ # Any Unicode letters/digits
See a regex demo.
C# usage:
string[] output = Regex.Matches(input, #"[\p{L}\p{N}]+").Cast<Match>().Select(x => x.Value).ToArray();
Yes, regex is indeed a good solution for this.
And in fact, to just match all standard words in the input sequence, this is all you need:
(\w+)
Let me quickly explain
\w matches any word character and is equivalent to [a-zA-Z0-9_] - matching a through z or A through Z or 0-9 or _, you might wanna go with [a-zA-Z0-9] instead to avoid that underscore.
Wrapping an expression in () means that you want to capture that part as a group.
The + means that you want sequences of 1 or more of the preceding characters.
Refer to a regular expression cheat sheet to see all the possibilities, such as
https://cheatography.com/davechild/cheat-sheets/regular-expressions/
Or any that you find online.
Also there are tools available to quickly test out your regular expressions, such as
https://regex101.com/ (quite well visualised matching)
or http://regexstorm.net/tester specifically for .NET
I'm trying to use regexes to match space-separated numbers.
I can't find a precise definition of \b ("word boundary").
I had assumed that -12 would be an "integer word" (matched by \b\-?\d+\b) but it appears that this does not work. I'd be grateful to know of ways of .
[I am using Java regexes in Java 1.6]
Example:
Pattern pattern = Pattern.compile("\\s*\\b\\-?\\d+\\s*");
String plus = " 12 ";
System.out.println(""+pattern.matcher(plus).matches());
String minus = " -12 ";
System.out.println(""+pattern.matcher(minus).matches());
pattern = Pattern.compile("\\s*\\-?\\d+\\s*");
System.out.println(""+pattern.matcher(minus).matches());
This returns:
true
false
true
A word boundary, in most regex dialects, is a position between \w and \W (non-word char), or at the beginning or end of a string if it begins or ends (respectively) with a word character ([0-9A-Za-z_]).
So, in the string "-12", it would match before the 1 or after the 2. The dash is not a word character.
In the course of learning regular expression, I was really stuck in the metacharacter which is \b. I indeed didn't comprehend its meaning while I was asking myself "what it is, what it is" repetitively. After some attempts by using the website, I watch out the pink vertical dashes at the every beginning of words and at the end of words. I got it its meaning well at that time. It's now exactly word(\w)-boundary.
My view is merely to immensely understanding-oriented. Logic behind of it should be examined from another answers.
A word boundary can occur in one of three positions:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
Word characters are alpha-numeric; a minus sign is not.
Taken from Regex Tutorial.
I would like to explain Alan Moore's answer
A word boundary is a position that is either preceded by a word character and not followed by one or followed by a word character and not preceded by one.
Suppose I have a string "This is a cat, and she's awesome", and I want to replace all occurrences of the letter 'a' only if this letter ('a') exists at the "Boundary of a word",
In other words: the letter a inside 'cat' should not be replaced.
So I'll perform regex (in Python) as
re.sub(r"\ba","e", myString.strip()) //replace a with e
Therefore,
Input; Output
This is a cat and she's awesome
This is e cat end she's ewesome
A word boundary is a position that is either preceded by a word character and not followed by one, or followed by a word character and not preceded by one.
I talk about what \b-style regex boundaries actually are here.
The short story is that they’re conditional. Their behavior depends on what they’re next to.
# same as using a \b before:
(?(?=\w) (?<!\w) | (?<!\W) )
# same as using a \b after:
(?(?<=\w) (?!\w) | (?!\W) )
Sometimes that isn’t what you want. See my other answer for elaboration.
I ran into an even worse problem when searching text for words like .NET, C++, C#, and C. You would think that computer programmers would know better than to name a language something that is hard to write regular expressions for.
Anyway, this is what I found out (summarized mostly from http://www.regular-expressions.info, which is a great site): In most flavors of regex, characters that are matched by the short-hand character class \w are the characters that are treated as word characters by word boundaries. Java is an exception. Java supports Unicode for \b but not for \w. (I'm sure there was a good reason for it at the time).
The \w stands for "word character". It always matches the ASCII characters [A-Za-z0-9_]. Notice the inclusion of the underscore and digits (but not dash!). In most flavors that support Unicode, \w includes many characters from other scripts. There is a lot of inconsistency about which characters are actually included. Letters and digits from alphabetic scripts and ideographs are generally included. Connector punctuation other than the underscore and numeric symbols that aren't digits may or may not be included. XML Schema and XPath even include all symbols in \w. But Java, JavaScript, and PCRE match only ASCII characters with \w.
Which is why Java-based regex searches for C++, C# or .NET (even when you remember to escape the period and pluses) are screwed by the \b.
Note: I'm not sure what to do about mistakes in text, like when someone doesn't put a space after a period at the end of a sentence. I allowed for it, but I'm not sure that it's necessarily the right thing to do.
Anyway, in Java, if you're searching text for the those weird-named languages, you need to replace the \b with before and after whitespace and punctuation designators. For example:
public static String grep(String regexp, String multiLineStringToSearch) {
String result = "";
String[] lines = multiLineStringToSearch.split("\\n");
Pattern pattern = Pattern.compile(regexp);
for (String line : lines) {
Matcher matcher = pattern.matcher(line);
if (matcher.find()) {
result = result + "\n" + line;
}
}
return result.trim();
}
Then in your test or main function:
String beforeWord = "(\\s|\\.|\\,|\\!|\\?|\\(|\\)|\\'|\\\"|^)";
String afterWord = "(\\s|\\.|\\,|\\!|\\?|\\(|\\)|\\'|\\\"|$)";
text = "Programming in C, (C++) C#, Java, and .NET.";
System.out.println("text="+text);
// Here is where Java word boundaries do not work correctly on "cutesy" computer language names.
System.out.println("Bad word boundary can't find because of Java: grep with word boundary for .NET="+ grep("\\b\\.NET\\b", text));
System.out.println("Should find: grep exactly for .NET="+ grep(beforeWord+"\\.NET"+afterWord, text));
System.out.println("Bad word boundary can't find because of Java: grep with word boundary for C#="+ grep("\\bC#\\b", text));
System.out.println("Should find: grep exactly for C#="+ grep("C#"+afterWord, text));
System.out.println("Bad word boundary can't find because of Java:grep with word boundary for C++="+ grep("\\bC\\+\\+\\b", text));
System.out.println("Should find: grep exactly for C++="+ grep(beforeWord+"C\\+\\+"+afterWord, text));
System.out.println("Should find: grep with word boundary for Java="+ grep("\\bJava\\b", text));
System.out.println("Should find: grep for case-insensitive java="+ grep("?i)\\bjava\\b", text));
System.out.println("Should find: grep with word boundary for C="+ grep("\\bC\\b", text)); // Works Ok for this example, but see below
// Because of the stupid too-short cutsey name, searches find stuff it shouldn't.
text = "Worked on C&O (Chesapeake and Ohio) Canal when I was younger; more recently developed in Lisp.";
System.out.println("text="+text);
System.out.println("Bad word boundary because of C name: grep with word boundary for C="+ grep("\\bC\\b", text));
System.out.println("Should be blank: grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));
// Make sure the first and last cases work OK.
text = "C is a language that should have been named differently.";
System.out.println("text="+text);
System.out.println("grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));
text = "One language that should have been named differently is C";
System.out.println("text="+text);
System.out.println("grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));
//Make sure we don't get false positives
text = "The letter 'c' can be hard as in Cat, or soft as in Cindy. Computer languages should not require disambiguation (e.g. Ruby, Python vs. Fortran, Hadoop)";
System.out.println("text="+text);
System.out.println("Should be blank: grep exactly for C="+ grep(beforeWord+"C"+afterWord, text));
P.S. My thanks to http://regexpal.com/ without whom the regex world would be very miserable!
Check out the documentation on boundary conditions:
http://java.sun.com/docs/books/tutorial/essential/regex/bounds.html
Check out this sample:
public static void main(final String[] args)
{
String x = "I found the value -12 in my string.";
System.err.println(Arrays.toString(x.split("\\b-?\\d+\\b")));
}
When you print it out, notice that the output is this:
[I found the value -, in my string.]
This means that the "-" character is not being picked up as being on the boundary of a word because it's not considered a word character. Looks like #brianary kinda beat me to the punch, so he gets an up-vote.
Reference: Mastering Regular Expressions (Jeffrey E.F. Friedl) - O'Reilly
\b is equivalent to (?<!\w)(?=\w)|(?<=\w)(?!\w)
Word boundary \b is used where one word should be a word character and another one a non-word character.
Regular Expression for negative number should be
--?\b\d+\b
check working DEMO
I believe that your problem is due to the fact that - is not a word character. Thus, the word boundary will match after the -, and so will not capture it. Word boundaries match before the first and after the last word characters in a string, as well as any place where before it is a word character or non-word character, and after it is the opposite. Also note that word boundary is a zero-width match.
One possible alternative is
(?:(?:^|\s)-?)\d+\b
This will match any numbers starting with a space character and an optional dash, and ending at a word boundary. It will also match a number starting at the beginning of the string.
when you use \\b(\\w+)+\\b that means exact match with a word containing only word characters ([a-zA-Z0-9])
in your case for example setting \\b at the begining of regex will accept -12(with space) but again it won't accept -12(without space)
for reference to support my words: https://docs.oracle.com/javase/tutorial/essential/regex/bounds.html
I think it's the boundary (i.e. character following) of the last match or the beginning or end of the string.
I'm looking for a regular expression to extract a string from a file name
eg if filename format is "anythingatallanylength_123_TESTNAME.docx", I'm interested in extracting "TESTNAME" ... probably fixed length of 8. (btw, 123 can be any three digit number)
I think I can use regex match ...
".*_[0-9][0-9][0-9]_[A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z].docx$"
However this matches the whole thing. How can I just get "TESTNAME"?
Thanks
Use parenthesis to match a specific piece of the whole regex.
You can also use the curly braces to specify counts of matching characters, and \d for [0-9].
In C#:
var myRegex = new Regex(#"*._\d{3}_([A-Za-z]{8})\.docx$");
Now "TESTNAME" or whatever your 8 letter piece is will be found in the captures collection of your regex after using it.
Also note, there will be a performance overhead for look-ahead and look-behind, as presented in some other solutions.
You can use a look-behind and a look-ahead to check parts without matching them:
(?<=_[0-9]{3}_)[A-Z]{8}(?=\.docx$)
Note that this is case-sensitive, you may want to use other character classes and/or quantifiers to fit your exact pattern.
In your file name format "anythingatallanylength_123_TESTNAME.docx", the pattern you are trying to match is a string before .docx and the underscore _. Keeping the thing in mind that any _ before doesn't get matched I came up with following solution.
Regex: (?<=_)[A-Za-z]*(?=\.docx$)
Flags used:
g global search
m multi-line search.
Explanation:
(?<=_) checks if there is an underscore before the file name.
(?=\.docx$) checks for extension at the end.
[A-Za-z]* checks the required match.
Regex101 Demo
Thanks to #Lucero #noob #JamesFaix I came up with ...
#"(?<=.*[0-9]{3})[A-Z]{8}(?=.docx$)"
So a look behind (in brackets, starting with ?<=) for anything (ie zero or more any char (denoted by "." ) followed by an underscore, followed by thee numerics, followed by underscore. Thats the end of the look behind. Now to match what I need (eight letters). Finally, the look ahead (in brackets, starting with ?=), which is the .docx
Nice work, fellas. Thunderbirds are go.
I am new to regex, I'm programming an advanced profanity filter for a commenting feature (in C#). Just to save time, I know that all filters can be fooled, no matter how good they are, you don't have to tell me that. I'm just trying to make it a bit more advanced than basic word replacement. I've split the task into several separate approaches and this is one of them.
What I need is a specific piece of regex, that catches strings such as these:
s_h_i_t
s h i t
S<>H<>I<>T
s_/h_/i_/t
s***h***i***t
you get the idea.
I guess what I'm looking for is a regex that says "one or more characters that are not alphanumeric". This should include both spaces and all special characters that you can type on a standard (western) keyboard. If possible, it should also include line breaks, so it would catch things like
s
h
i
t
There should always be at least one of the characters present, to avoid likely false positives such as in
Finish it.
This will of course mean that things like
sh_it
will not be caught, but as I said, it doesn't matter, it doesn't have to be perfect. All I need is the regex, I can do the splitting of words and inserting the regex myself. I have the RegexOptions.IgnoreCase option set in my C# code, so character case in the actual word is not an issue. Also, this regex shouldn't worry about "leetspeek", i.e. some of the actual letters of the word being replaced by other characters:
sh1t
I have a different approach that deals with that.
Thank you in advance for your help.
Lets see if this regex works for you:
/\w(?:_|\W)+/
Alright, HamZa's answer worked. However I ran into a programmatic problem while working on the solution. When I was replacing just the words, I always knew the length of the word. So I knew exactly how many asterisks to replace it with. If I'm matching shit, I know I need to put 4 asterisks. But if I'm matching s[^a-z0-9]+h[^a-z0-9]+[^a-z0-9]+i[^a-z0-9]+t, I might catch s#h#i#t or I may catch s------h------i--------t. In both cases the length of the matched text will differ wildly from that of the pattern. How can I get the actual length of the matched string?
\bs[\W_]*h[\W_]*i[\W_]*t[\W_]*(?!\w)
matches characters between letters that aren't word characters or character _ or whitespace characters (also new line breaks)
\b (word boundrary) ensures that Finish it won't match
(?!\w) ensures that sh ituuu wont match, you may want to remove/modify that, as s_hittt will not match as well. \bs[\W_]*h[\W_]*i[\W_]*t+[\W_]*(?!\w) will match the word with repeated last character
modification \bs[\W_]*h[\W_]*i[\W_]*t[\W_]*?(?!\w) will make the match of last character class not greedy and in sh it&&& only sh it will match
\bs[\W\d_]*h[\W\d_]*i[\W\d_]*t+[\W\d_]*?(?!\w) will match sh1i444t (digits between characters)
EDIT:
(?!\w) is a negative lookahead. It basicly checks if your match is followed by a word character (word characters are [A-z09_]). It has a length of 0, which means it won't be included in the match. If you want to catch words like "shi*tface" you'll have to remove it.
( http://www.regular-expressions.info/lookaround.html )
A word booundrary [/b] matches a place where word starts or ends, it's length is 0, which means that it matches between characters
[\W] is a negative character class, I think it's equal to [^a-zA-Z0-9_] or [^\w]
You want to match words where each letter is separated with the identical non-word char(s).
You can use
\b\p{L}(?=([\W_]+))(?:\1\p{L})+\b
See the regex demo. (I added (?!\n) to make the regex work for each line as if it were a separate string.) Details:
\b - word boundary
\p{L} - a letter
(?=([\W_]+)) - a positive lookahead that matches a location that is immediately followed with any non-word or _ char (captured into Group 1)
(?:\1\p{L})+ - one or more repetitions of a sequence of the same char captured into Group 1 and a letter
\b - word boundary.
To check if there is such a pattern in a string, you can use
var HasSpamWords = Regex.IsMatch(text, #"\b\p{L}(?=([\W_]+))(?:\1\p{L})+\b");
To return all occurrences in a string, you can use
var results = Regex.Matches(text, #"\b\p{L}(?=([\W_]+))(?:\1\p{L})+\b")
.Cast<Match>()
.Select(x => x.Value)
.ToList();
See the C# demo.
Getting the length of each string is easy if you get Match.Length and use .Select(x => x.Length). If you need to get the length of the string with all special chars removed, simply use .Select(x => x.Value.Count(c => char.IsLetter(c))) (see this C# demo).