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Regex JS no output in JS [duplicate]

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.

Regular Expression for Blocking a character in begining

I am facing an issue with a regular expression while trying to block any string which has minus(-) in the beginning of some white listed characters.
^(?!-.*$).([a-zA-Z0-9-:#\\,()\\/\\.]+)$
It is blocking minus(-) at place and allowing it any where in the character sequence but this regex is not working if the passed string is single character.
For e.g A or 9 etc.
Please help me out with this or give me a good regex to do the task.

Your pattern requires at least 2 chars in the input string because there is a dot after the first lookahead and then a character class follows that has + after it (that is, at least 1 occurrence must be present in the string).
So, you need to remove the dot. Also, you do not need to escape any special char inside a character class. Besides, to avoid matching strings atarting with - a mere (?!-) will suffice, no need adding .*$ there. You may use
^(?!-)[a-zA-Z0-9:#,()/.-]+$
See the regex demo. Remember to escape / if used in a regex literal notation in JavaScript, there is no need to escape it in a constructor notation or in a Java regex pattern.
Details
^ - start of a string
(?!-) - cannot start with -
[a-zA-Z0-9:#,()/.-]+ - 1 or more ASCII letters, digits and special chars defined in the character class (:, #, ,, (, ), /, ., -)
$ - end of string.

If i understand correctly, and you don't want a minus at the beginning, does ^[^-].* work as a regex for you? Java's "matches" would return false if it starts with minus

There is a method in a String class that provides you exactly what you are asking for - it's a startsWith() method - you could use this method in your code like this (you can translate it as "If the given String doesn't start with -, doSomething, in other case do the else part, that can contain some code or might be empty if you want nothing to be done if the given String starts with - ") :
if(!(yourString.startsWith("-"))) {
doSomething()
} else {
doNothingOrProvideAnyInformationAboutWrongInput()
}

I think that it can help you.
^(?!-).*[a-zA-Z0-9-:#\\,()\/\\.]+$

Regex match with '\' slash and replace with '\\'?

I was converting normal string in to latex format.So i was created the latex code match and replace the \ single slash into \\ double slash.why the i need it Refer this link.I tried Below code :
function test(){
var tex="$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$";
var tex_form = tex.replace("/[\\\/\\\\\.\\\\]/g", "\\");
document.getElementById('demo').innerHTML=tex_form;//nothing get
}
test();
<p id="demo"></p>
Not getting any output data.But the match in this link
i wish to need replace the \ into \\

There are these issues:
The string literal has no backslashes;
The regular expression is not a regular expression;
The class in the intended regular expression cannot match sequences, only single characters;
The replacement would not add backslashes, only replace with them.
Here you find the details on each point:
1. How to Encode Backslashes in String Literals
Your tex variable has no backslashes. This is because a backslash in a string literal is not taken as a literal backslash, but as an escape for interpreting the character that follows it.
When you have "$$\left...", then the \l means "literal l", and so the content of your variable will be:
$$left...
As an l does not need to be escaped, the backslash is completely unnecessary, and these two assignments result in the same string value:
var tex="$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$";
var tex="$$left[ x=left({{11}over{2}}+{{sqrt{3271}}over{2,3^{{{3}over{2} $$";
To bring the point home, this will also represent the same value:
var tex="\$\$\l\e\f\t\[\ \x\=\l\e\f\t\(\{\{\1\1\}\o\v\e\r\{\2\}\}\+\{\{\s\q\r\t\{\3\2\7\1\}\}\o\v\e\r\{\2\,\3\^\{\{\{\3\}\o\v\e\r\{\2\}\ \$\$";
If you really want to have literal backslashes in your content (which I understand you do, as this is about LaTeX), then you need to escape each of those backslashes... with a backslash:
var tex="$$\\left[ x=\\left({{11}\\over{2}}+{{\\sqrt{3271}}\\over{2\\,3^{{{3}\\over{2} $$";
Now the content of your tex variable will be this string:
$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$
2. How to Code Regular Expression Literals
You are passing a string literal to the first argument of replace, while you really intend to pass a regular expression literal. You should leave out the quotes for that to happen. The / are the delimiters of a regular expression literal, not quotes:
/[\\\/\\\\\.\\\\]/g
This should not be wrapped in quotes. JavaScript understands the / delimiters as denoting a regular expression literal, including the optional modifiers at the end (like g here).
3. Classes are sets of single characters
This regular expression has unnecessary characters. The class [...] should list all individual characters you want to match. Currently you have these characters (after resolving the escapes):
\
/
\
\
.
\
\
It is overkill to have the backslash represented 5 times. Also, in JavaScript the forward slash and dot do not need to be escaped when occurring in a class. So the above regular expression is equivalent to this one:
/[\\/.]/g
Maybe this is, or is not, what you intended to match. To match several sequences of characters, you could use the | operator. This is just an example:
/\\\\|\\\/|\\\./g
... but I don't think you need this.
4. How to actually prefix with backslashes
It seems strange to me that you would want to replace a point or forward slash with a backslash. Probably you want to prefix those with a backslash. In that case make a capture group (with parentheses) and refer to it with $1 in this replace:
tex.replace(/([\\/.])/g, "\\$1");
Note again, that in the replacement string there is only one literal backslash, as the first one is an escape (see point 1 above).

why the i need it
As the question you link to says, the \ character has special meaning inside a JavaScript string literal. It represents an escape sequence.
Not getting any output data.But the match in this link
The escape sequence is processed when the string literal is parsed by the JavaScript compiler.
By the time you apply your regular expression to them, they have been consumed. The slash characters only exist in your source code, not in your data.
If you want to put a slash character in your string, then you need to write the escape sequence for it (the \\) in the source code. You can't add them back in with JavaScript afterwards.

Not sure if I understood the problem, but try this code:
var tex_form = tex.replace("/(\\)/g","\\\\");.
You need to use '(' ')' instead of '['']' to get a match for output.

unable to parse - in Regular expression in Javascript

I am a bit new to the regular expressions in Javascript.
I am trying to write a function called parseRegExpression()
which parses the attributes passed and generates a key/value pairs
It works fine with the input:
"iconType:plus;iconPosition:bottom;"
But it is not able to parse the input:
"type:'date';locale:'en-US';"
Basically the - sign is being ignored. The code is at:
http://jsfiddle.net/visibleinvisibly/ZSS5G/
The Regular Expression key value pair is as below
/[a-z|A-Z|-]*\s*:\s*[a-z|A-Z|'|"|:|-|_|\/|\.|0-9]*\s*;|[a-z|A-Z|-]*\s*:\s*[a-z|A-Z|'|"|:|-|_|\/|\.|0-9]*\s*$/gi;

There are a few problems:
A | inside a character class means a literal | character, not an alternation.
A . inside a character class means a literal . character, so there's no need to escape it.
A - as the first or last character inside a character class means a literal - character, otherwise it means a character range.
There's no need to use [a-zA-Z] when you use the case-insensitive modifier (i); [a-z] is enough.
The only difference between your alterations is the last bit; this can be simplified significantly by just limiting your alternation to that part which is different.
This should be equivalent to your original pattern:
/[a-z-]*\s*:\s*[a-z0-9'":_\/.-]*\s*(?:;|$)/gi

You can avoid the regex:
var test1 = "iconType:plus;iconPosition:bottom;";
var test2 = "type:'date';locale:'en-US';";
function toto(str) {
var result = new Array();
var temp = str.split(';');
for (i=0; i<temp.length-1; i++) {
result[i] = temp[i].split(':',1);
}
return result;
}
console.log(toto(test1));
console.log(toto(test2));

Inside a character set atom [...] the pipe char | is just a regular char and doesn't mean "or".
A character set atom lists characters or ranges you want to accept (or exclude if the character set starts with ^) and "or" is implicit.
You can use a backslash in a character set if you need to include/exclude a close bracket ], the ^ sign, the dash - that is used for ranges, the backslash \ itself, an unprintable character or if you want to use a non-ASCII unicode char specifying the code instead of literally.
Regular expression syntax however also lets you to avoid backslash-escaping in a character set atom by placing the character in a position where it cannot have the special meaning... for example a dash - as first or last in the set (it cannot mean a range there).
Note also that if you need to be able to match as values quoted strings, including backslash escaping, the regular expression is more complex, for example
'(?:[^'\\]|\\.)*'|"(?:[^"\\]|\\.)*"
matches a single-quoted or double-quoted string including backslash escaping, the meaning being:
A single quote '
Zero or more of either:
Any char except the single quote ' or the backslash \
A pair composed of a backslash \ followed by any char
A single quote '
or the same with double quotes " instead.
Note that the groups have been delimited with (?:...) instead of plain (...) to avoid capture

It doesn't match hyphens because it interpreting |-| as a range that starts at | and ends at |. (I would have expected that to be treated as a syntax error, but there you have it. It works the same in every regex flavor I've tried, too.)
Have a look at this regex:
/(?:^|;)([a-z-]*)\s*:\s*([a-z'":_\/.0-9-]*)\s*(?=;|$)/ig
As suggested by the other responders, I collapsed it to one alternative, removed the unneeded pipes, and escaped the hyphen by moving it to the end. I also anchored it at the beginning as well as the end. Or anchored it as well as I can, anyway. I used a lookahead to match the trailing semicolon so it will still be there when the next match starts. It's far from foolproof, but it should work okay as long as the input is well formed.

Replace regular expressions in your code as follow:
regExpKeyValuePair = /[-a-z]*\s*:\s*[-a-z'":_\/.0-9]*\s*;|[-a-z]*\s*:\s*[-a-z'":-_\/.0-9]*\s*$/gi;
regExpKey = /[-a-z]*/gi;
regExpValue = /[-a-z:_\/.0-9]*/gi;
You don't need escape . inside [].
No need to put | between elements [].
Because you are using /i flag, [A-Z] is not needed.
- should be at the beginning or at the end.

Writing a Javascript regex that includes special reserved characters

I'm writing a function that takes a prospective filename and validates it in order to ensure that no system disallowed characters are in the filename. These are the disallowed characters: / \ | * ? " < >
I could obviously just use string.indexOf() to search for each special char one by one, but that's a lot longer than it would be to just use string.search() using a regular expression to find any of those characters in the filename.
The problem is that most of these characters are considered to be part of describing a regular expression, so I'm unsure how to include those characters as actually being part of the regex itself. For example, the / character in a Javascript regex tells Javascript that it is the beginning or end of the regex. How would one write a JS regex that functionally behaves like so: filename.search(\ OR / OR | OR * OR ? OR " OR < OR >)

Put your stuff in a character class like so:
[/\\|*?"<>]
You're gonna have to escape the backslash, but the other characters lose their special meaning. Also, RegExp's test() method is more appropriate than String.search in this case.
filenameIsInvalid = /[/\\|*?"<>]/.test(filename);

Include a backslash before the special characters [\^$.|?*+(){}, for instance, like \$
You can also search for a character by specified ASCII/ANSI value. Use \xFF where FF are 2 hexadecimal digits. Here is a hex table reference. http://www.asciitable.com/ Here is a regex reference http://www.regular-expressions.info/reference.html

The correct syntax of the regex is:
/^[^\/\\|\*\?"<>]+$/
The [^ will match anything, but anything that is matched in the [^] group will return the match as null. So to check for validation is to match against null.
Demo: jsFiddle.
Demo #2: Comparing against null.
The first string is valid; the second is invalid, hence null.
But obviously, you need to escape regex characters that are used in the matching. To escape a character that is used for regex needs to have a backslash before the character, e.g. \*, \/, \$, \?.

You'll need to escape the special characters. In javascript this is done by using the \ (backslash) character.
I'd recommend however using something like xregexp which will handle the escaping for you if you wish to match a string literal (something that is lacking in javascript's native regex support).