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.
I'm having trouble with a certain RegEx replacement string for later use in Javascript.
We have quite a bit of text that was stored in a rather odd format that we aren't allowed to fix.
But we do need to find all the "network path" strings inside it, following these rules:
A. The matches always start with 2 backslashes.
B. The matching characters should stop as soon as it hits a first occurrence of any 1 of these:
A < character
A space
A line feed
A carriage return
A & character
A literal "\r" or "\n" string (but only if occurring at end of line)
We "almost" have it working with /\\\\[^ &<\s]*/gi as shown in this RegEx Tester page:
https://regex101.com/r/T4cDOL/5
Even if we get it working, the RegEx has to be even futher "escape escaped" before putting on
our Javascript code, but that's also not working as expected.
From your example, it seems you literally have a backslash followed by an n and a backslash followed by an r (as opposed to a newline or carriage return), which means you can't only use a negated character class (since you need to handle a sequence of two characters). I'd use a positive lookahead to know where to stop, so I can use an alternation for that part.
You haven't said what parts of those strings should match, so I've had to guess a bit, but here's my best guess (with useful input from Niet the Dark Absol):
const rex = /\\\\.*?(?=[ &<\r\n]|\\[rn](?:$| ))/gmi;
That says:
Match starting with \\
Take everything prior to the lookahead (non-greedy)
Lookahead: An alternation of:
A space, &, <, carriage return (\r, character 13), or a newline (\n, character 10); or
A backslash followed by r or n if that's either at the end of a line or followed by a space (so we get the \nancy but not the \n after it).
Updated regex101
You might want to have more characters than just a space after the \r/\n. If so, make it a character class (and/or use \s for "whitespace" if that applies):
const rex = /\\\\.*?(?=[ &<\r\n]|\\[rn](?:$|[ others]))/gmi;
// −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^^^^^^^^^
Using this bit of code trims out hidden characters like carriage returns and linefeeds with nothing using javascript just fine:
value = value.replace(/[\r\n]*/g, "");
but when the code actually contains \r\n text what do I do to trim it without affecting r's and n's in my content? I've tried this code:
value = value.replace(/[\\r\\n]+/g, "");
on this bit of text:
{"client":{"werdfasreasfsd":"asdfRasdfas\r\nMCwwDQYJKoZIhvcNAQEBBQADGw......
I end up with this:
{"cliet":{"wedfaseasfsd":"asdfRasdfasMCwwDQYJKoZIhvcNAQEBBQADGw......
Side note: It leaves the upper case versions of R and N alone because I didn't include the /i flag at the end and thats ok in this case.
What do I do to just remove \r\n text found in the string?
If you want to match literal \r and literal \n then you should use the following:
value = value.replace(/(?:\\[rn])+/g, "");
You might think that matching literal \r and \n with [\\r\\n] is the right way to do it and it is a bit confusing but it won't work and here is why:
Remember that in character classes, each single character represents a single letter or symbol, it doesn't represent a sequence of characters, it is just a set of characters.
So the character class [\\r\\n] actually matches the literal characters \, r and n as separate letters and not as sequences.
Edit: If you want to replace all carriage returns \r, newlines \n and also literal \r and '\n` then you could use:
value = value.replace(/(?:\\[rn]|[\r\n]+)+/g, "");
About (?:) it means a non-capturing group, because by default when you put something into a usual group () then it gets captured into a numbered variable that you can use elsewhere inside the regular expression itself, or latter in the matches array.
(?:) prevents capturing the value and causes less overhead than (), for more info see this article.
To just remove them, this seems to work for me:
value = value.replace(/[\r\n]/g, "");
You don't need the * after the character set because the g flag solves that for you.
Note, this will remove all \r or \n chars whether they are in this exact sequence or not.
Working demo of this option: http://jsfiddle.net/jfriend00/57GtJ/
If you want to remove these characters only when in this exact sequence (e.g. only when a \r is directly followed by a \n, you could use this:
value = value.replace(/\r\n/g, "");
Working demo of this option: http://jsfiddle.net/jfriend00/Ta3sn/
If you have text with a lot of \r\n and want to save all of them try this one
value.replace(/(?:\\[rn]|[\r\n])/g,"<br>")
http://jsfiddle.net/57GtJ/63/
I do not know what I am doing wrong. I have this string that I want to replace
<?xml version="1.0" encoding="utf-8" ?>
<Sections>
<Section>
I am using regex to replace everything including <Section>, and leave the rest untouched.
arrayValues[index].replace("/[([.,\n,\s])*<Section>]/", "---");
What is wrong with my regex? Doesn't this mean repalce every character, including new line and spaces, up to and including <Section> with ---?
First of all, you need to remove the quotes around your regex—if they're there, the argument won't be processed as a regex. JavaScript will see it as a string (because it is a string) and try to match it literally.
Now that that's taken care of, we can simplify your regex a bit:
arrayValues[index].replace(/[\s\S]*?<Section>/, "---");
[\s\S] gets around JavaScript's lack of an s flag (a handy option supported by most languages that enables . to match newlines). \s does match newlines (even without an s flag specified), so the character class [\s\S] tells the regex engine to match:
\s - a whitespace character, which could be a newline
OR
\S - a non-whitespace character
So you can think of [\s\S] as matching . (any character except a newline) or the literal \n (a newline). See Javascript regex multiline flag doesn't work for more.
? is used to make the initial [\s\S]* match non-greedy, so the regex engine will stop once it hits the first occurrence of <Section>.
arrayValues[index].replace("/[([.,\n,\s])*<Section>]/", "---");
What is wrong with my regex?
It's no regex, it's string literal. A string would be converted to a regex, but yours would then include the slashes. Use a regex literal instead:
arrayValues[index].replace(/[\S\s]*<Section>/, "---");
Also, you have too many unnecessary characters in it. The [] around the whole thing build a character class, which is not what you want. The capturing group () just wraps a character class which can be repeated itself. And a dot . inside a character class does match a literal dot, instead of all characters.
How do you split a long piece of text into separate lines? Why does this return line1 twice?
/^(.*?)$/mg.exec('line1\r\nline2\r\n');
["line1", "line1"]
I turned on the multi-line modifier to make ^ and $ match beginning and end of lines. I also turned on the global modifier to capture all lines.
I wish to use a regex split and not String.split because I'll be dealing with both Linux \n and Windows \r\n line endings.
arrayOfLines = lineString.match(/[^\r\n]+/g);
As Tim said, it is both the entire match and capture. It appears regex.exec(string) returns on finding the first match regardless of global modifier, wheras string.match(regex) is honouring global.
Use
result = subject.split(/\r?\n/);
Your regex returns line1 twice because line1 is both the entire match and the contents of the first capturing group.
I am assuming following constitute newlines
\r followed by \n
\n followed by \r
\n present alone
\r present alone
Please Use
var re=/\r\n|\n\r|\n|\r/g;
arrayofLines=lineString.replace(re,"\n").split("\n");
for an array of all Lines including the empty ones.
OR
Please Use
arrayOfLines = lineString.match(/[^\r\n]+/g);
For an array of non empty Lines
Even simpler regex that handles all line ending combinations, even mixed in the same file, and removes empty lines as well:
var lines = text.split(/[\r\n]+/g);
With whitespace trimming:
var lines = text.trim().split(/\s*[\r\n]+\s*/g);
Unicode Compliant Line Splitting
Unicode® Technical Standard #18 defines what constitutes line boundaries. That same section also gives a regular expression to match all line boundaries. Using that regex, we can define the following JS function that splits a given string at any line boundary (preserving empty lines as well as leading and trailing whitespace):
const splitLines = s => s.split(/\r\n|(?!\r\n)[\n-\r\x85\u2028\u2029]/)
I don't understand why the negative look-ahead part ((?!\r\n)) is necessary, but that is what is suggested in the Unicode document 🤷♂️.
The above document recommends to define a regular expression meta-character for matching all line ending characters and sequences. Perl has \R for that. Unfortunately, JavaScript does not include such a meta-character. Alas, I could not even find a TC39 proposal for that.
First replace all \r\n with \n, then String.split.
http://jsfiddle.net/uq55en5o/
var lines = text.match(/^.*((\r\n|\n|\r)|$)/gm);
I have done something like this. Above link is my fiddle.