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+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
I have a string. The end is different, such as index.php?test=1&list=UL or index.php?list=UL&more=1. The one thing I'm looking for is &list=.
How can I match it, whether it's in the middle of the string or it's at the end? So far I've got [&|\?]list=.*?([&|$]), but the ([&|$]) part doesn't actually work; I'm trying to use that to match either & or the end of the string, but the end of the string part doesn't work, so this pattern matches the second example but not the first.
Use:
/(&|\?)list=.*?(&|$)/
Note that when you use a bracket expression, every character within it (with some exceptions) is going to be interpreted literally. In other words, [&|$] matches the characters &, |, and $.
In short
Any zero-width assertions inside [...] lose their meaning of a zero-width assertion. [\b] does not match a word boundary (it matches a backspace, or, in POSIX, \ or b), [$] matches a literal $ char, [^] is either an error or, as in ECMAScript regex flavor, any char. Same with \z, \Z, \A anchors.
You may solve the problem using any of the below patterns:
[&?]list=([^&]*)
[&?]list=(.*?)(?=&|$)
[&?]list=(.*?)(?![^&])
If you need to check for the "absolute", unambiguous string end anchor, you need to remember that is various regex flavors, it is expressed with different constructs:
[&?]list=(.*?)(?=&|$) - OK for ECMA regex (JavaScript, default C++ `std::regex`)
[&?]list=(.*?)(?=&|\z) - OK for .NET, Go, Onigmo (Ruby), Perl, PCRE (PHP, base R), Boost, ICU (R `stringr`), Java/Andorid
[&?]list=(.*?)(?=&|\Z) - OK for Python
Matching between a char sequence and a single char or end of string (current scenario)
The .*?([YOUR_SINGLE_CHAR_DELIMITER(S)]|$) pattern (suggested by João Silva) is rather inefficient since the regex engine checks for the patterns that appear to the right of the lazy dot pattern first, and only if they do not match does it "expand" the lazy dot pattern.
In these cases it is recommended to use negated character class (or bracket expression in the POSIX talk):
[&?]list=([^&]*)
See demo. Details
[&?] - a positive character class matching either & or ? (note the relationships between chars/char ranges in a character class are OR relationships)
list= - a substring, char sequence
([^&]*) - Capturing group #1: zero or more (*) chars other than & ([^&]), as many as possible
Checking for the trailing single char delimiter presence without returning it or end of string
Most regex flavors (including JavaScript beginning with ECMAScript 2018) support lookarounds, constructs that only return true or false if there patterns match or not. They are crucial in case consecutive matches that may start and end with the same char are expected (see the original pattern, it may match a string starting and ending with &). Although it is not expected in a query string, it is a common scenario.
In that case, you can use two approaches:
A positive lookahead with an alternation containing positive character class: (?=[SINGLE_CHAR_DELIMITER(S)]|$)
A negative lookahead with just a negative character class: (?![^SINGLE_CHAR_DELIMITER(S)])
The negative lookahead solution is a bit more efficient because it does not contain an alternation group that adds complexity to matching procedure. The OP solution would look like
[&?]list=(.*?)(?=&|$)
or
[&?]list=(.*?)(?![^&])
See this regex demo and another one here.
Certainly, in case the trailing delimiters are multichar sequences, only a positive lookahead solution will work since [^yes] does not negate a sequence of chars, but the chars inside the class (i.e. [^yes] matches any char but y, e and s).
I'm having trouble with a certain RegEx replacement string for later use in Javascript.
We have quite a bit of text that was stored in a rather odd format that we aren't allowed to fix.
But we do need to find all the "network path" strings inside it, following these rules:
A. The matches always start with 2 backslashes.
B. The matching characters should stop as soon as it hits a first occurrence of any 1 of these:
A < character
A space
A line feed
A carriage return
A & character
A literal "\r" or "\n" string (but only if occurring at end of line)
We "almost" have it working with /\\\\[^ &<\s]*/gi as shown in this RegEx Tester page:
https://regex101.com/r/T4cDOL/5
Even if we get it working, the RegEx has to be even futher "escape escaped" before putting on
our Javascript code, but that's also not working as expected.
From your example, it seems you literally have a backslash followed by an n and a backslash followed by an r (as opposed to a newline or carriage return), which means you can't only use a negated character class (since you need to handle a sequence of two characters). I'd use a positive lookahead to know where to stop, so I can use an alternation for that part.
You haven't said what parts of those strings should match, so I've had to guess a bit, but here's my best guess (with useful input from Niet the Dark Absol):
const rex = /\\\\.*?(?=[ &<\r\n]|\\[rn](?:$| ))/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](?:$|[ others]))/gmi;
// −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^^^^^^^^^
What is the difference between "\\w+#\\w+[.]\\w+" and "^\\w+#\\w+[.]\\w+$"? I have tried to google for it but no luck.
^ means "Match the start of the string" (more exactly, the position before the first character in the string, so it does not match an actual character).
$ means "Match the end of the string" (the position after the last character in the string).
Both are called anchors and ensure that the entire string is matched instead of just a substring.
So in your example, the first regex will report a match on email#address.com.uk, but the matched text will be email#address.com, probably not what you expected. The second regex will simply fail.
Be careful, as some regex implementations implicitly anchor the regex at the start/end of the string (for example Java's .matches(), if you're using that).
If the multiline option is set (using the (?m) flag, for example, or by doing Pattern.compile("^\\w+#\\w+[.]\\w+$", Pattern.MULTILINE)), then ^ and $ also match at the start and end of a line.
Try the Javadoc:
http://download.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
^ and $ match the beginnings/endings of a line (without consuming them)
Tried to search for /\,$/ online, but coudnt find anything.
I have:
coords = coords.replace(/\,$/, "");
Im guessing it returns coords string index number. What I have to search online for this, so I can learn more?
/\,$/ finds the comma character (,) at the end of a string (denoted by the $) and replaces it with empty (""). You sometimes see this in regex code aiming to clean up excerpts of text.
It's a regular expression to remove a trailing comma.
That thing is a Regular Expression, also known as regex or regexp. It is a way to "match" strings using some rules. If you want to learn how to use it in JavaScript, read the Mozilla Developer Network page about RegExp.
By the way, regular expressions are also available on most languages and in some tools. It is a very useful thing to learn.
That's a regular expression that finds a comma at the end of a string. That code removes the comma.
// defines a JavaScript regular expression, used to match a pattern within a string.
\,$ is the pattern
In this case \, translates to ,. A backslash is used to escape special characters, but in this case, it's not necessary. An example where it would be necessary would be to remove trailing periods. If you tried to do that with /.$/ the period here has a different meaning; it is used as a wildcard to match [almost] any character (aside for some newlines). So in this case to match on "." (period character) you would have to escape the wildcard (/\.$/).
When $ is placed at the end of the pattern, it means only look at the end of the string. This means that you can't mistakingly find a comma anywhere in the middle of the string (e.g., not after help in help, me,), only at the end (trailing). It also speeds of the regular expression search considerably. If you wanted to match on characters only at the beginning of the string, you would start off the pattern with a carat (^), for instance /^,/ would find a comma at the start of a string if one existed.
It's also important to note that you're only removing one comma, whereas if you use the plus (+) after the comma, you'd be replacing one or more: /,+$/.
Without the +; trailing commas,, becomes trailing commas,
With the +; no trailing comma,, becomes no trailing comma