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.
So I'm relatively n00bish at regular expressions, and doing a little practicing.
I'm playing with a dog-simple "deobfucator" that just looks for [dot] or (dot) or [at] or (at). Case-insensitive, and with or w/out any number of spaces before or after the match(s).
This is for the usual: someemail [AT] domain (dot) com type of thing. I want to obviously turn it into someemail#domain.com.
The RegEx I've come up with does the matching fine, but now I want to replace with either a . or a # depending on the match.
i.e.
I want the group matching the "dot" group to replace it with the literal ., and the group matching the "at" group with the literal #.
I know I could just write 2 different (almost identical) RegEx's and run it through both, but for the sake of education, I'm trying to see if I can do it all in one RegEx?
Here's the RegEx I came up with (probably not the smallest possible, which I'd also be interested in seeing):
+(\[|\()(dot)(\)|\]) +| +(\[|\()(at)(\)|\]) +
NOTE: before each + there's an empty space, for matching spaces.
What I'm looking for is what I would use to do the replacement(s) properly?
Update: Sorry all, forgot to add which language I was working with for this. In this case, I'm using a clipboard utility that can run RegEx's on it's input (whatever gets copied to the clipboard), and the engine it uses is C#/VB.NET. Ultimate goal for this little project is to just be able to copy an "obfuscated" email address or URL, and run the RegEx on it so that it's set on the clipboard in it's "unobfuscated" state.
That said, I do tend to use RegEx's on many different languages, so converting them between languages generally isn't an issue.
.NET regex does not support conditional replacement patterns.
for the sake of education, I'm trying to see if I can do it all in one RegEx?
There are other regex engines that allow conditional replacement logic in a single regex replacement operation with conditional replacement patterns.
There are 3 engines that support this type of replacements: JGsoft V2, Boost, and PCRE2.
For conditionals to work in Boost, you need to pass regex_constants::format_all to regex_replace. For them to work in PCRE2, you need to pass PCRE2_SUBSTITUTE_EXTENDED to pcre2_substitute.
In PCRE2:
${1:+matched:unmatched} where 1 is a number between 1 and 99 referencing a numbered capturing group. If your regex contains named capturing groups then you can reference them in a conditional by their name: ${name:+matched:unmatched}.
If you want a literal colon in the matched part, then you need to escape it with a backslash. If you want a literal closing curly brace anywhere in the conditional, then you need to escape that with a backslash too. Plus signs have no special meaning beyond the :+ that starts the conditional, so they don't need to be escaped.
Also, see The Boost-Specific Format Sequences:
When specifying the format_all flag to regex_replace(), the escape sequences recognized are the same as those above for format_perl. In addition, conditional expressions of the following form are recognized:
?Ntrue-expression:false-expression
where N is a decimal digit representing a sub-match. If the corresponding sub-match participated in the full match, then the substitution is true-expression. Otherwise, it is false-expression. In this mode, you can use parens () for grouping. If you want a literal paren, you must escape it as \(.
In Boost replacement patterns, literal ( and ) must be escaped.
The syntax for JGsoft V2 replacement string conditionals is the same as that in the C++ Boost library.
So, your regex can be contracted to ( +)[[(](?:(dot)|(at))[])]( +):
( +) - Group 1: one or more spaces
[[(] - a [ or (
(?:(dot)|(at)) - Either (Group 2) a dot substring or (Group 3) an at substring
[])] - a ) or ]
( +) - Group 4: one or more spaces
And replace with $1(?{3}.:#)$4:
$1 - Group 1 value,
(?{3}.:#) - if Group 3 matched, replace with ., else with #
$4 - Group 4 value.
This is available in Notepad++:
If you are using Java, try replaceAll method from String class.
And finally you need to normalize it with white spaces:
- Pure Java - String after = before.trim().replaceAll("\\s+", " ");
- Pure Java - String after = before.replaceAll("\\s{2,}", " ").trim();
- Apache commons lang3 - String after = StringUtils.normalizeSpace(String str);
- ...
I have the following:
https://www.example.com/my-suburl/sub-dept/xx-xxxx-xx-yyyyyy/
Im trying to find the 'yyyyy' in the url so far I have:
(.*)\/sub-dept\/(.*[^\/])\/([^\/]*)$
Which matches on:
https://www.example.com/my-suburl
and
xx-xxxx-xx-yyyyyy
However like i say I need the 'yyyyy' specific match
NON-C#-BASED SOLUTION
If xx are numbers in the actual strings, just use
\d+(?=\/$)
Or else, use
[^-\/]*(?=\/?$)
See Demo 1 and Demo 2
Note that in JS, there is no look-behind, thus, if you must check if /sub-dept/ is in front of the substring you need, you will have to rely on capturing group mechanism:
\/sub-dept\/[^\/]*-([^-\/]*)\/?
See yet another demo
ORIGINAL ANSWER
Here is a regex you can use
(?<=/sub-dept/[^/]*-)[^/-]*(?=/$)
See demo
The regex matches a substring that contains 0 or more characters other than a / or - that is...
(?<=/sub-dept/[^/]*-) preceded with /sub-dept/ followed by 1 or more characters other than / and then a hyphen
(?=/$) - is followed by a / symbol right at the end of the string.
Or, there is a non-regex way: split the string by /, get the last part and split by -. Here is an example (without error/null checking for demo sake):
var result = text.Trim('/').Split('/').LastOrDefault().Split('-').LastOrDefault();
I am new to RegEx and thus have a question on RegEx. I am writing my code in C# and need to come up with a regex to find matching strings.
The possible combination of strings i get are,
XYZF44DT508755
ABZF44DT508755
PQZF44DT508755
So what i need to check is whether the string starts with XY or AB or PQ.
I came up with this one and it doesn't work.
^((XY|AB|PQ).){2}
Note: I don't want to use regular string StartsWith()
UPDATE:
Now if i want to try a new matching condition like this -
If string starts with "XY" or "AB" or "PQ" and 3rd character is "Z" and 4th character is "F"
How to write the RegEx for that?
You can modify you expression to the following and use the IsMatch() method.
Regex.IsMatch(input, "^(?:XY|AB|PQ)")
The outer capturing group in conjuction with . (any single character) is trying to match a third character and then repeat the sequence twice because of the range quantifier {2} ...
According to your updated edit, you can simply place "ZF" after the grouping construct.
Regex.IsMatch(input, "^(?:XY|AB|PQ)ZF")
You want to test for just ^(XY|AB|PQ). Your RegEx means: Search for either XY, AB or PQ, then a random character, and repeat the whole sequence twice, for example "XYKPQL" would match your RegEx.
This is a screenshot of the matches on regex101:
^ forces the start of line,
(...) creates a matching group and
XY|AB|PQ matches either XY, AB or PQ.
If you want the next two characters to be ZF, just append ZF to the RegEx so it becomes ^(XY|AB|PQ)ZF.
Check out regex101, a great way to test your RegExes.
You were on the right track. ^(XY|AB|PQ) should match your string correctly.
The problem with ^((XY|AB|PQ).){2} is following the entire group with {2}. This means exactly 2 occurrences. That would be 2 occurrences of your first 2 characters, plus . (any single character), meaning this would match strings like XY_AB_. The _ could be anything.
It may have been your intention with the . to match a larger string. In this case you might try something along the lines of ^((XY|AB|PQ)\w*). The \w* will match 0 or more occurrences of "word characters", so this should match all of XYZF44DT508755 up to a space, line break, punctuation, etc., and not just the XY at the beginning.
There are some good tools out there for understanding regexes, one of my favorites is debuggex.
UPDATE
To answer your updated question:
If string starts with "XY" or "AB" or "PQ" and 3rd character is "Z" and 4th character is "F"
The regex would be (assuming you want to match the entire "word").
^((XY|AB|PQ)ZF\w*)
Debuggex Demo
I need a help with regular expression as I do not have good knowledge in it.
I have regular expression as:
Regex myregex = new Regex("testValue=\"(.+?)\"");
What does (.+?) indicate?
The string it matches is "testValue=123e4567" and returns 123e4567 as output.
Now I need help in regular expression to match a string "<helpMe>123e4567</helpMe>" where I need 123e4567 as output. How do I write a regular expression for it?
This means:
( Begin captured group
. Match any character
+ One or more times
? Non-greedy quantifier
) End captured group
In the case of your regex, the non-greedy quantifier ? means that your captured group will begin after the first double-quote, and then end immediately before the very next double-quote it encounters. If it were greedy (without the ?), the group would extend to the very last double-quote it encounters on that line (i.e., "greedily" consuming as much of the line as possible).
For your "helpMe" example, you'd want this regex:
<helpMe>(.+?)</helpMe>
Given this string:
<div>Something<helpMe>ABCDE</helpMe></div>
You'd get this match:
ABCDE
The value of the non-greedy quantifier is evident in this variation:
Regex: <helpMe>(.+)</helpMe>
String: <div>Something<helpMe>ABCDE</helpMe><helpMe>FGHIJ</helpMe></div>
The greedy capture would look like this:
ABCDE</helpMe><helpMe>FGHIJ
There are some useful interactive tools to play with these variations:
Regex Tester
Regex Pal
Ken Redler has a great answer regarding your first question. For the second question try:
<(helpMe)>(.*?)</\1>
Using the back reference \1 you can find values between the set of matching tags. The first group finds the tag name, the second group matches the content itself, and the \1 back reference re-uses the first group's match (in this case the tag name).
Also, in C# you can use named groups, like: <(helpMe)>(?<value>.*?)</\1> where now match.Groups["value"].Value contains your value.
What does (.+?) indicate?
It means match any character (.) one or more times (+?)
A simple regex to match your second string would be
<helpMe>([a-z0-9]+)<\/helpMe>
This will match any character of a-z and any digit inside <helpme> and </helpMe>.
The pharanteses are used to capture a group. This is useful if you need to reference the value inside this group later.