Related
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.
Simple regex question. I have a string on the following format:
this is a [sample] string with [some] special words. [another one]
What is the regular expression to extract the words within the square brackets, ie.
sample
some
another one
Note: In my use case, brackets cannot be nested.
You can use the following regex globally:
\[(.*?)\]
Explanation:
\[ : [ is a meta char and needs to be escaped if you want to match it literally.
(.*?) : match everything in a non-greedy way and capture it.
\] : ] is a meta char and needs to be escaped if you want to match it literally.
(?<=\[).+?(?=\])
Will capture content without brackets
(?<=\[) - positive lookbehind for [
.*? - non greedy match for the content
(?=\]) - positive lookahead for ]
EDIT: for nested brackets the below regex should work:
(\[(?:\[??[^\[]*?\]))
This should work out ok:
\[([^]]+)\]
Can brackets be nested?
If not: \[([^]]+)\] matches one item, including square brackets. Backreference \1 will contain the item to be match. If your regex flavor supports lookaround, use
(?<=\[)[^]]+(?=\])
This will only match the item inside brackets.
To match a substring between the first [ and last ], you may use
\[.*\] # Including open/close brackets
\[(.*)\] # Excluding open/close brackets (using a capturing group)
(?<=\[).*(?=\]) # Excluding open/close brackets (using lookarounds)
See a regex demo and a regex demo #2.
Use the following expressions to match strings between the closest square brackets:
Including the brackets:
\[[^][]*] - PCRE, Python re/regex, .NET, Golang, POSIX (grep, sed, bash)
\[[^\][]*] - ECMAScript (JavaScript, C++ std::regex, VBA RegExp)
\[[^\]\[]*] - Java, ICU regex
\[[^\]\[]*\] - Onigmo (Ruby, requires escaping of brackets everywhere)
Excluding the brackets:
(?<=\[)[^][]*(?=]) - PCRE, Python re/regex, .NET (C#, etc.), JGSoft Software
\[([^][]*)] - Bash, Golang - capture the contents between the square brackets with a pair of unescaped parentheses, also see below
\[([^\][]*)] - JavaScript, C++ std::regex, VBA RegExp
(?<=\[)[^\]\[]*(?=]) - Java regex, ICU (R stringr)
(?<=\[)[^\]\[]*(?=\]) - Onigmo (Ruby, requires escaping of brackets everywhere)
NOTE: * matches 0 or more characters, use + to match 1 or more to avoid empty string matches in the resulting list/array.
Whenever both lookaround support is available, the above solutions rely on them to exclude the leading/trailing open/close bracket. Otherwise, rely on capturing groups (links to most common solutions in some languages have been provided).
If you need to match nested parentheses, you may see the solutions in the Regular expression to match balanced parentheses thread and replace the round brackets with the square ones to get the necessary functionality. You should use capturing groups to access the contents with open/close bracket excluded:
\[((?:[^][]++|(?R))*)] - PHP PCRE
\[((?>[^][]+|(?<o>)\[|(?<-o>]))*)] - .NET demo
\[(?:[^\]\[]++|(\g<0>))*\] - Onigmo (Ruby) demo
If you do not want to include the brackets in the match, here's the regex: (?<=\[).*?(?=\])
Let's break it down
The . matches any character except for line terminators. The ?= is a positive lookahead. A positive lookahead finds a string when a certain string comes after it. The ?<= is a positive lookbehind. A positive lookbehind finds a string when a certain string precedes it. To quote this,
Look ahead positive (?=)
Find expression A where expression B follows:
A(?=B)
Look behind positive (?<=)
Find expression A where expression B
precedes:
(?<=B)A
The Alternative
If your regex engine does not support lookaheads and lookbehinds, then you can use the regex \[(.*?)\] to capture the innards of the brackets in a group and then you can manipulate the group as necessary.
How does this regex work?
The parentheses capture the characters in a group. The .*? gets all of the characters between the brackets (except for line terminators, unless you have the s flag enabled) in a way that is not greedy.
Just in case, you might have had unbalanced brackets, you can likely design some expression with recursion similar to,
\[(([^\]\[]+)|(?R))*+\]
which of course, it would relate to the language or RegEx engine that you might be using.
RegEx Demo 1
Other than that,
\[([^\]\[\r\n]*)\]
RegEx Demo 2
or,
(?<=\[)[^\]\[\r\n]*(?=\])
RegEx Demo 3
are good options to explore.
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
Test
const regex = /\[([^\]\[\r\n]*)\]/gm;
const str = `This is a [sample] string with [some] special words. [another one]
This is a [sample string with [some special words. [another one
This is a [sample[sample]] string with [[some][some]] special words. [[another one]]`;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
Source
Regular expression to match balanced parentheses
(?<=\[).*?(?=\]) works good as per explanation given above. Here's a Python example:
import re
str = "Pagination.go('formPagination_bottom',2,'Page',true,'1',null,'2013')"
re.search('(?<=\[).*?(?=\])', str).group()
"'formPagination_bottom',2,'Page',true,'1',null,'2013'"
The #Tim Pietzcker's answer here
(?<=\[)[^]]+(?=\])
is almost the one I've been looking for. But there is one issue that some legacy browsers can fail on positive lookbehind.
So I had to made my day by myself :). I manged to write this:
/([^[]+(?=]))/g
Maybe it will help someone.
console.log("this is a [sample] string with [some] special words. [another one]".match(/([^[]+(?=]))/g));
if you want fillter only small alphabet letter between square bracket a-z
(\[[a-z]*\])
if you want small and caps letter a-zA-Z
(\[[a-zA-Z]*\])
if you want small caps and number letter a-zA-Z0-9
(\[[a-zA-Z0-9]*\])
if you want everything between square bracket
if you want text , number and symbols
(\[.*\])
This code will extract the content between square brackets and parentheses
(?:(?<=\().+?(?=\))|(?<=\[).+?(?=\]))
(?: non capturing group
(?<=\().+?(?=\)) positive lookbehind and lookahead to extract the text between parentheses
| or
(?<=\[).+?(?=\]) positive lookbehind and lookahead to extract the text between square brackets
In R, try:
x <- 'foo[bar]baz'
str_replace(x, ".*?\\[(.*?)\\].*", "\\1")
[1] "bar"
([[][a-z \s]+[]])
Above should work given the following explaination
characters within square brackets[] defines characte class which means pattern should match atleast one charcater mentioned within square brackets
\s specifies a space
+ means atleast one of the character mentioned previously to +.
I needed including newlines and including the brackets
\[[\s\S]+\]
If someone wants to match and select a string containing one or more dots inside square brackets like "[fu.bar]" use the following:
(?<=\[)(\w+\.\w+.*?)(?=\])
Regex Tester
I have this regex that removes all non-ascii characters from a string including all smart quotes:
str.replace(/[\u{0080}-\u{FFFF}]/gu,"");
But I need to keep the Smart quotes
The regex for removing Smart single quotes is: [\u2018\u2019\u201A\u201B\u2032\u2035] and for Smart double quotes is: [\u201C\u201D\u201E\u201F\u2033\u2036].
I need a combined regex that that removes all non-ASCII ([\u{0080}-\u{FFFF}]) except smart quotes ([\u2018\u2019\u201A\u201B\u2032\u2035] or [\u201C\u201D\u201E\u201F\u2033\u2036]).
Note that you need to use the \u{XXXX} notation in the regex with u modifier, and to build the regex you need you need to put the character class with exceptions into a negative lookahead placed right before your more generic pattern:
/(?![\u{2018}\u{2019}\u{201A}\u{201B}\u{2032}\u{2035}\u{201C}\u{201D}\u{201E}\u{201F}\u{2033}\u{2036}])[\u{0080}-\u{FFFF}]/gu
See the regex demo
Note that some chars in the Unicode table go one after another, so we may shorten the pattern using ranges:
/(?![\u{2018}-\u{201F}\u{2032}\u{2033}\u{2035}\u{2036}])[\u{0080}-\u{FFFF}]/gu
See this demo.
Instead of matching the non-ascii, match the ascii + the characters you need, and negate the expression. Example:
str.replace(/[^\x00-\x7F\u2018\u2019\u201A\u201B\u2032\u2035\u201C\u201D\u201E\u201F\u2033\u2036]/gu,"");
I want to build a localization application for my javascript (pebble.js) application. It should find all Strings between l._(" and ") or ') or ", or ',.
So for example I have the the line
console.log(l._("This is a Test") + l._('%# times %# equals %#', 2, 4, (2*4)));
With the Swift application I should get an Array like this:
["This is a Test", "%# times %# equals %#"]
Right now I have no clue how I should manage it. Should I use a Regex, NSScanner or should I split the strings?
Thanks!
I'm not familiar with NSScanner, but with regexp the solution would be quite easy, with the assumption, that a string has the same delimiter at both ends. (I.e., you don't have stuff like l._("Hello world').) I think that wouldn't be valid syntax in JavaScript, so let's assume that is the case.
Also, let's assume that the strings don't contain any escaped quotes (of the same kind that is used as delimiter), i.e. there are no such strings: l._("Hello \" world).
Now you could use the following two regexps to find strings delimited by double quotes and those delimited by single quotes:
l._\("((?:[^"])*)" -- for double quotes
l._\('((?:[^'])*)' -- for single quotes
Then, you have to run these two regexps on your input, and get the result of the first capturing group for each match. (I'm not sure how exactly swift uses regexps, but note that in many implementation, capturing group #0 is usually the whole match, and capturing group #1 is what is located between the first pair of parenthesis -- you need the latter.)
Also, note that you don't have to care what comes after the closing quote: whether it's a parenthesis, or a comma, it's the same for you, as you only have to look up to the quote.
EDIT: Corrected the regexp (repetition should be inside capturing group).
EDIT2: If we allow escaped quotes, than you can use this regexp for the single quoted case:
l._\('((?:\\'|[^'])*)'. And a similar one for the double quoted.
You can play around with the regex on this link:
https://regex101.com/r/tY1jB5/1
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