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.
Related
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
For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
I'm using a regex pattern that I got from regexlib to validate relative urls. On their site you can test the pattern to make sure it fits your needs. Everything works great on their site, as soon as I use the pattern in mine I get the error message:
Invalid range in character class
I know that this error usually means that a hyphen is mistakenly being used to represent a range and is not properly escaped. But in this case since it works on their site I'm confused why it's not working on mine.
var urlRegex = new RegExp('^(?:(?:\.\./)|/)?(?:\w(?:[\w`~!$=;\-\+\.\^\(\)\|\{\}\[\]]|(?:%\d\d))*\w?)?(?:/\w(?:[\w`~!$=;\-\+\.\^\(\)\|\{\}\[\]]|(?:%\d\d))*\w?)*(?:\?[^#]+)?(?:#[a-z0-9]\w*)?$', 'g');
NOTE:
If you're going to test the regex from their site (using the link above) be sure to change the Regex Engine dropdown to Client-side Engine and the Engine dropdown to Javascript.
Either put - at the end or beginning of the character class or use two backslashes to do a regex escape within string
since you are using string you need to use two backslashes for each special characters..
NOTE
Check out this answer on SO which explains when to use single or double backslashes to escape special characters
There is no reason to use RegExp constructor here. Just use RegExp literal:
var urlRegex = /^(?:(?:\.\.\/)|\/)?(?:\w(?:[\w`~!$=;\-\+\.\^\(\)\|\{\}\[\]]|(?:%\d\d))*\w?)?(?:\/\w(?:[\w`~!$=;\-\+\.\^\(\)\|\{\}\[\]]|(?:%\d\d))*\w?)*(?:\?[^#]+)?(?:#[a-z0-9]\w*)?$/g;
^ ^ ^ ^ ^
Inside RegExp literal, you just write the regex naturally, except for /, which now needs escaping, since / is used as delimiter in the RegExp literal.
In character class, ^ has special meaning at the beginning of the character class, - has special meaning in between 2 characters, and \ has special meaning, which is to escape other characters (mainly ^, -, [, ] and \) and also to specify shorthand character classes (\d, \s, \w, ...). [, ] are used as delimiters for character class, so they also have special meaning. (Actually, in JavaScript, only ] has special meaning, and you can specify [ without escaping inside character class). Other than those 5 character listed above, other characters (unless involved in an escape sequence with \) doesn't have any special meaning.
You can reduce the number of escaping \ with the information above. For ^, unless it is the only character in the character class, you can put it away from the beginning of the character class. For -, you can put it at the end of the character class.
var urlRegex = /^(?:(?:\.\.\/)|\/)?(?:\w(?:[\w`~!$=;+.^()|{}\[\]-]|(?:%\d\d))*\w?)?(?:\/\w(?:[\w`~!$=;+.^()|{}\[\]-]|(?:%\d\d))*\w?)*(?:\?[^#]+)?(?:#[a-z0-9]\w*)?$/g;
What was changed:
[\w`~!$=;\-\+\.\^\(\)\|\{\}\[\]]
[\w`~!$=;+.^()|{}\[\]-]
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).