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.
The regex allows chars that are: alphanumeric, space, '-', '_', '&', '()' and '/'
this is the expression
[\s\/\)\(\w&-]
I have tested this in various online testers and know it works, I just can't get it to work correctly in code. I get sysntax errors with anything I try.. any suggestions?
var programProductRegex = new RegExp([\s\/\)\(\w&-]);
You can use the regular expression syntax:
var programProductRegex = /[\s\/\)\(\w&-]/;
You use forward slashes to delimit the regex pattern.
If you use the RegExp object constructor you need to pass in a string. Because backslashes are special escape characters inside JavaScript strings and they're also escape characters in regular expressions, you need to use two backslashes to do a regex escape inside a string. The equivalent code using a string would then be:
var programProductRegex = new RegExp("[\\s\\/\\)\\(\\w&-]");
All the backslashes that were in the original regular expression need to be escaped in the string to be correctly interpreted as backslashes.
Of course the first option is better. The constructor is helpful when you obtain a string from somewhere and want to make a regular expression out of it.
var programProductRegex = new RegExp(userInput);
If you are using a String and want to escape characters like (, you need to write \\( (meaning writing backslash, then the opening parenthesis => escaping it).
If you are using the RegExp object, you only need one backslash for each character (like \()
Enclose your regex with delimiters:
var programProductRegex = /[\s\/)(\w&-]/;
Say I have this:
console.log(new RegExp('.git'));
console.log(new RegExp('scripts/npm'));
the results are:
/.git/
/scripts\/npm/
my question is - why does it escape the slash in scripts/npm, but it does not escape the . in .git? What is the rhyme and reason to that?
Note, in this case, the regex strings are being passed from the command line, so I need to convert them to regex using RegExp.
An unescaped / denotes the beginning and end of a regular expression. When you pass in a string containing / into the constructor, of course that / is part of the regular expression, not a symbol denoting the beginning or end.
The . is something else entirely, and has nothing to RE delimiters, so it's left as-is.
Note that if you want the regular expression to match a literal dot (rather than any character), you need to double-escape it when using the constructor:
console.log(new RegExp('\\.git'));
When you write regex in JS you can initialize regex strings using two /. This is called regular expression literal initialization. More about it here.
For instance
let re = /(\w+)\s(\w+)/;
Now, for the question why does it append \ before /, it is simply due to the way RegExp processes the passed string literal. This prevents the passed string from becoming corrupt ensuring all passed characters are accounted.
Furthermore if you examine the object returned by the RegExp, we can see that the actual source attribute is set to scripts\\/npm. So, the first \ indicates literal significance of the second \. Whilst from regular expressions perspective \, it simply escapes the proceeding / to form regular expression literal notation.
I am trying to build a regexp from static text plus a variable in javascript. Obviously I am missing something very basic, see comments in code below. Help is very much appreciated:
var test_string = "goodweather";
// One regexp we just set:
var regexp1 = /goodweather/;
// The other regexp we built from a variable + static text:
var regexp_part = "good";
var regexp2 = "\/" + regexp_part + "weather\/";
// These alerts now show the 2 regexp are completely identical:
alert (regexp1);
alert (regexp2);
// But one works, the other doesn't ??
if (test_string.match(regexp1))
alert ("This is displayed.");
if (test_string.match(regexp2))
alert ("This is not displayed.");
First, the answer to the question:
The other answers are nearly correct, but fail to consider what happens when the text to be matched contains a literal backslash, (i.e. when: regexp_part contains a literal backslash). For example, what happens when regexp_part equals: "C:\Windows"? In this case the suggested methods do not work as expected (The resulting regex becomes: /C:\Windows/ where the \W is erroneously interpreted as a non-word character class). The correct solution is to first escape any backslashes in regexp_part (the needed regex is actually: /C:\\Windows/).
To illustrate the correct way of handling this, here is a function which takes a passed phrase and creates a regex with the phrase wrapped in \b word boundaries:
// Given a phrase, create a RegExp object with word boundaries.
function makeRegExp(phrase) {
// First escape any backslashes in the phrase string.
// i.e. replace each backslash with two backslashes.
phrase = phrase.replace(/\\/g, "\\\\");
// Wrap the escaped phrase with \b word boundaries.
var re_str = "\\b"+ phrase +"\\b";
// Create a new regex object with "g" and "i" flags set.
var re = new RegExp(re_str, "gi");
return re;
}
// Here is a condensed version of same function.
function makeRegExpShort(phrase) {
return new RegExp("\\b"+ phrase.replace(/\\/g, "\\\\") +"\\b", "gi");
}
To understand this in more depth, follows is a discussion...
In-depth discussion, or "What's up with all these backslashes!?"
JavaScript has two ways to create a RegExp object:
/pattern/flags - You can specify a RegExp Literal expression directly, where the pattern is delimited using a pair of forward slashes followed by any combination of the three pattern modifier flags: i.e. 'g' global, 'i' ignore-case, or 'm' multi-line. This type of regex cannot be created dynamically.
new RegExp("pattern", "flags") - You can create a RegExp object by calling the RegExp() constructor function and pass the pattern as a string (without forward slash delimiters) as the first parameter and the optional pattern modifier flags (also as a string) as the second (optional) parameter. This type of regex can be created dynamically.
The following example demonstrates creating a simple RegExp object using both of these two methods. Lets say we wish to match the word "apple". The regex pattern we need is simply: apple. Additionally, we wish to set all three modifier flags.
Example 1: Simple pattern having no special characters: apple
// A RegExp literal to match "apple" with all three flags set:
var re1 = /apple/gim;
// Create the same object using RegExp() constructor:
var re2 = new RegExp("apple", "gim");
Simple enough. However, there are significant differences between these two methods with regard to the handling of escaped characters. The regex literal syntax is quite handy because you only need to escape forward slashes - all other characters are passed directly to the regex engine unaltered. However, when using the RegExp constructor method, you pass the pattern as a string, and there are two levels of escaping to be considered; first is the interpretation of the string and the second is the interpretation of the regex engine. Several examples will illustrate these differences.
First lets consider a pattern which contains a single literal forward slash. Let's say we wish to match the text sequence: "and/or" in a case-insensitive manner. The needed pattern is: and/or.
Example 2: Pattern having one forward slash: and/or
// A RegExp literal to match "and/or":
var re3 = /and\/or/i;
// Create the same object using RegExp() :
var re4 = new RegExp("and/or", "i");
Note that with the regex literal syntax, the forward slash must be escaped (preceded with a single backslash) because with a regex literal, the forward slash has special meaning (it is a special metacharacter which is used to delimit the pattern). On the other hand, with the RegExp constructor syntax (which uses a string to store the pattern), the forward slash does NOT have any special meaning and does NOT need to be escaped.
Next lets consider a pattern which includes a special: \b word boundary regex metasequence. Say we wish to create a regex to match the word "apple" as a whole word only (so that it won't match "pineapple"). The pattern (as seen by the regex engine) needs to be: \bapple\b:
Example 3: Pattern having \b word boundaries: \bapple\b
// A RegExp literal to match the whole word "apple":
var re5 = /\bapple\b/;
// Create the same object using RegExp() constructor:
var re6 = new RegExp("\\bapple\\b");
In this case the backslash must be escaped when using the RegExp constructor method, because the pattern is stored in a string, and to get a literal backslash into a string, it must be escaped with another backslash. However, with a regex literal, there is no need to escape the backslash. (Remember that with a regex literal, the only special metacharacter is the forward slash.)
Backslash SOUP!
Things get even more interesting when we need to match a literal backslash. Let's say we want to match the text sequence: "C:\Program Files\JGsoft\RegexBuddy3\RegexBuddy.exe". The pattern to be processed by the regex engine needs to be: C:\\Program Files\\JGsoft\\RegexBuddy3\\RegexBuddy\.exe. (Note that the regex pattern to match a single backslash is \\ i.e. each must be escaped.) Here is how you create the needed RegExp object using the two JavaScript syntaxes
Example 4: Pattern to match literal back slashes:
// A RegExp literal to match the ultimate Windows regex debugger app:
var re7 = /C:\\Program Files\\JGsoft\\RegexBuddy3\\RegexBuddy\.exe/;
// Create the same object using RegExp() constructor:
var re8 = new RegExp(
"C:\\\\Program Files\\\\JGsoft\\\\RegexBuddy3\\\\RegexBuddy\\.exe");
This is why the /regex literal/ syntax is generally preferred over the new RegExp("pattern", "flags") method - it completely avoids the backslash soup that can frequently arise. However, when you need to dynamically create a regex, as the OP needs to here, you are forced to use the new RegExp() syntax and deal with the backslash soup. (Its really not that bad once you get your head wrapped 'round it.)
RegexBuddy to the rescue!
RegexBuddy is a Windows app that can help with this backslash soup problem - it understands the regex syntaxes and escaping requirements of many languages and will automatically add and remove backslashes as required when pasting to and from the application. Inside the application you compose and debug the regex in native regex format. Once the regex works correctly, you export it using one of the many "copy as..." options to get the needed syntax. Very handy!
You should use the RegExp constructor to accomplish this:
var regexp2 = new RegExp(regexp_part + "weather");
Here's a related question that might help.
The forward slashes are just Javascript syntax to enclose regular expresions in. If you use normal string as regex, you shouldn't include them as they will be matched against. Therefore you should just build the regex like that:
var regexp2 = regexp_part + "weather";
I would use :
var regexp2 = new RegExp(regexp_part+"weather");
Like you have done that does :
var regexp2 = "/goodweather/";
And after there is :
test_string.match("/goodweather/")
Wich use match with a string and not with the regex like you wanted :
test_string.match(/goodweather/)
While this solution may be overkill for this specific question, if you want to build RegExps programmatically, compose-regexp can come in handy.
This specific problem would be solved by using
import {sequence} from 'compose-regexp'
const weatherify = x => sequence(x, /weather/)
Strings are escaped, so
weatherify('.')
returns
/\.weather/
But it can also accept RegExps
weatherify(/./u)
returns
/.weather/u
compose-regexp supports the whole range of RegExps features, and let one build RegExps from sub-parts, which helps with code reuse and testability.
I've got this regular expression for validating phone numbers
^(\+?|(\(\+?[0-9]{1,3}\))|)([ 0-9.//-]|\([ 0-9.//-]+\))+((x|X|((e|E)(x|X)(t|T)))([ 0-9.//-]|\([ 0-9.//-]+\)))?$
I dugged it out from my C#/vb library and now i want to translate it into javascript. But it has syntax error (i suspect it is something due to the // characters). my attempt:
$IsPhone = function (input) {
var regex = new window.RegExp("^$|^(\+?|(\(\+?[0-9]{1,3}\))|)([ 0-9.//-]|\([ 0-9.//-]+\))+((x|X|((e|E)(x|X)(t|T)))([ 0-9.//-]|\([ 0-9.//-]+\)))?$", "");
return regex.test(input.trim());
};
alert($IsPhone("asd"));
Your problem has nothing to do with comments. You're just mixing up the two different ways of creating RegExp objects.
When you create a RegExp object in JavaScript code, you either write it as a string literal which you pass to a RegExp constructor, or as a regex literal. Because string literals support backslash-escape sequences like \n and \", any actual backslash in the string has to be escaped, too. So, whenever you need to escape a regex metacharacter like ( or +, you have to use two backslashes, like so:
var r0 = "^$|^(\\+?|(\\(\\+?[0-9]{1,3}\\))|)([ 0-9./-]|\\([ 0-9./-]+\\))+((x|X|((e|E)(x|X)(t|T)))([ 0-9./-]|\\([ 0-9./-]+\\)))?$";
var regex0 = new RegExp(r0, "");
The forward-slash has no special meaning, either to regexes or strings. The only reason you ever have to escape forward-slashes is because they're used as the delimiter for regex literals. You use backslashes to escape the forward-slashes just like you do with regex metacharacters like \( or \+, or the backslash itself: \\. Here's the regex-literal version of your regex:
var regex1 = /^$|^(\+?|(\(\+?[0-9]{1,3}\))|)([ 0-9.\/-]|\([ 0-9.\/-]+\))+((x|X|((e|E)(x|X)(t|T)))([ 0-9.\/-]|\([ 0-9.\/-]+\)))?$/;
from Errors translating regex from .NET to javascript
The backslash character in JavaScript
strings is an escape character, so the
backslashes you have in your string
are escaping the next character for
the string, not for the regular
expression. So right near the
beginning, in your "^(+?, the
backslash there just escapes the + for
the string (which it doesn't need),
and what the regexp sees is just a raw
+ with nothing to repeat. Hence the error.