Regular Expressions- Match Anything

Question

How do I make an expression to match absolutely anything (including whitespaces)?
Example:

Regex: I bought _____ sheep.

Matches: I bought sheep. I bought a sheep. I bought five sheep.

I tried using (.*), but that doesn't seem to be working.

It's not working because there is two spaces between "bought" and "sheep". so `I bought sheep` is wrong and `I bought sheep` is correct. — , Oct 01 '20 at 05:19
`(?s:.)` - the [inline modifier group](https://www.regular-expressions.info/modifiers.html) match any char including line break chars. In your case, it would be something like this: `(?s:.*?)`. Taken from the [answer](https://stackoverflow.com/a/33312193/15164646) by [Wiktor Stribiżew](https://stackoverflow.com/users/3832970/wiktor-stribi%c5%bcew). — Dmitriy Zub, Dec 27 '21 at 12:33

Tim Pietzcker · Answer 1 · 2013-08-28T07:26:09.420

435

Normally the dot matches any character except newlines.

So if .* isn't working, set the "dot matches newlines, too" option (or use (?s).*).

If you're using JavaScript, which doesn't have a "dotall" option, try [\s\S]*. This means "match any number of characters that are either whitespace or non-whitespace" - effectively "match any string".

Another option that only works for JavaScript (and is not recognized by any other regex flavor) is [^]* which also matches any string. But [\s\S]* seems to be more widely used, perhaps because it's more portable.

edited Aug 28 '13 at 07:26

answered Jul 15 '11 at 19:09

Tim Pietzcker

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`.*` doesn't match `\n` but it maches a string that contains only `\n` because it matches 0 character. – Toto Aug 28 '13 at 09:45
1

if using Javascript don't forget to slash the slash when setting a variable to this pattern eg: var pattern = "[\\s\\S]*"; – Astra Bear Aug 24 '16 at 23:41
2

the `(?s).*` works for me matched everything including new line – Gujarat Santana May 28 '18 at 06:00
1

There is a PROBLEM with using /[\s\S]*/ . If you use it in your code and then comment out such code that causes a syntax error because the end of the pattern is taken to mean the end of the comment. You then need to remove that pattern from the commented-out code to make it work. But then if you ever un-comment it again it will not work any more, or not work like it used to, and should. – Panu Logic Oct 24 '18 at 17:10
This would be the case with any regex ending in an asterisk. Of course it's easy to avoid by either adding something after `*` (as in `/.*(?:)/`) or using the regex constructor (`var foo = new RegExp(".*")`). – Tim Pietzcker Oct 24 '18 at 19:11
I pretty much never want to match 0 of something, so I just use [\s\S]+ instead. – Asa Stallard Dec 06 '18 at 05:19
In 2018, the [dotall flag](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/dotAll) was added to the specification. Here we are approaching 2020 and Firefox still hasn't implemented it. Give them some [heat](https://bugzilla.mozilla.org/show_bug.cgi?id=1361856). – Lonnie Best Dec 16 '19 at 11:14
Also, it looks like JavaScript actually has dotall now https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/dotAll ? So you could also just – bersling Apr 23 '21 at 08:57
@Tim Pietzcker good one! ((?s).*) works perfectly!!! Thanks! – Sahelanthropus Jun 29 '22 at 09:54

score 246 · Answer 2 · edited Jan 29 '18 at 10:18

246

(.*?) matches anything - I've been using it for years.

edited Jan 29 '18 at 10:18

Adam Arold

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answered Mar 06 '15 at 18:34

jdixon2614

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but does that match a newline? – ineedahero Oct 20 '16 at 14:45
I used this on Sublime Text and it worked perfectly. Thanks! `("title":".*?")` – SCabralO Jul 06 '17 at 14:14
13

In Javascript, use `[\s\S]*` or `[^]*`. – Jeff Huijsmans Mar 21 '18 at 09:53
I don't know but every time I use this expression, I feel guilty, for not making a specific expression for my use case. If say, `w+` isn't enough, I end up using `.+`. Luckily hasn't come back to bite me yet. – Tushar Oct 01 '18 at 12:26
Works for me using Python – suffa Jul 05 '19 at 21:19
2

This doesn't seem to include new lines or trailing full stops. I found this slight alteration hits everything (?s).* – Mitchb Feb 27 '20 at 08:49
That matches 0 or more of anything included in the "." – Sam P Feb 22 '21 at 13:28
This definitely doesn't work in Python - in fact, it will do the exact opposite. The `?` makes the expression non-greedy, so it will match *nothing*. Thus: `re.search('(.*)', 'abc').group(1)` -> `'abc'` and `re.search('(.*?)', 'abc').group(1)` -> `''`. In Python, the correct syntax is `(?s)(.*)` (or `re.S` / `re.DOTALL` can be passed in the `flags` argument). – ekhumoro Aug 20 '21 at 13:29

Tyler · Answer 3 · 2017-03-10T18:32:50.380

130

Choose & memorize 1 of the following!!! :)

[\s\S]*
[\w\W]*
[\d\D]*

Explanation:

\s: whitespace \S: not whitespace

\w: word \W: not word

\d: digit \D: not digit

(You can exchange the * for + if you want 1 or MORE characters [instead of 0 or more]).

BONUS EDIT:

If you want to match everything on a single line, you can use this:

[^\n]+

Explanation:

^: not

\n: linebreak

+: for 1 character or more

edited Mar 10 '17 at 18:32

answered Feb 11 '16 at 23:38

Tyler

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score 30 · Answer 4 · answered Jul 15 '11 at 19:08

30

Try this:

I bought (.* )?sheep

or even

I bought .*sheep

answered Jul 15 '11 at 19:08

Mike Mozhaev

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Danny · Answer 5 · 2020-12-13T09:33:48.840

27

/.*/ works great if there are no line breaks. If it has to match line breaks, here are some solutions:

Solution	Description
`/.*/s`	`/s` (dot all flag) makes `.` (wildcard character) match anything, including line breaks. Throw in an `*` (asterisk), and it will match everything. Read more.
`/[\s\S]*/`	`\s` (whitespace metacharacter) will match any whitespace character (space; tab; line break; ...), and `\S` (opposite of `\s`) will match anything that is not a whitespace character. `*` (asterisk) will match all occurrences of the character set (Encapsulated by `[]`). Read more.

edited Dec 13 '20 at 09:33

answered Jul 12 '20 at 10:01

Danny

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does it match no chars? Want match one or more. – mathtick Dec 18 '20 at 11:15
Yes, the solutions I posted above will also match nothing/empty string. If you want to match at least one character, simply replace `*` with `+`. – Danny Dec 18 '20 at 22:48
1

`/[\s\S]*/` works to me to match anything ignore break lines – Marco Mesen Aug 12 '21 at 21:37
2

You have saved me almost another 48 hours of search – John Max Aug 21 '22 at 22:02

score 12 · Answer 6 · answered May 13 '19 at 03:22

12

Because . Find a single character, except newline or line terminator.

So, to match anything, You can use like this: (.|\n)*?

Hope it helps!

answered May 13 '19 at 03:22

Sang Huynh

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Maybe I am a bit ignorant, but to my knowledge the `?` makes the preceding quantifier lazy, causing it to match as few characters as possible (in the case, zero!) – Danny Dec 16 '20 at 17:13
Also, this will only work with LF (`\n`) line endings, but not CR (`\r`) or CRLF (`\r\n`) line endings. – Danny Dec 16 '20 at 17:17
nope, not in dart it doesn't – MwBakker Jul 06 '23 at 18:17

score 7 · Answer 7 · edited Dec 12 '20 at 21:55

7

If you're using JavaScript, ES2018 added the /s (dotAll) flag. With the /s flag, the dot . will match any character, including a newline.

console.log("line_1\nline_2".match(/.+/s))

Note: It's not supported by all browsers yet.

edited Dec 12 '20 at 21:55

Danny

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answered Sep 16 '19 at 02:47

Cuong Le Ngoc

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score 7 · Answer 8 · answered Jul 15 '11 at 19:11

7

Use .*, and make sure you are using your implementations' equivalent of single-line so you will match on line endings.

There is a great explanation here -> http://www.regular-expressions.info/dot.html

answered Jul 15 '11 at 19:11

Josh

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in dart it doesn't – MwBakker Jul 06 '23 at 18:16
1

Good point @MwBakker. If you could lend me your time machine, I’ll travel back to July 11, 2011 and tell myself to wait a few months for Dart to be released. – Josh Jul 31 '23 at 18:20

oryxfea · Answer 9 · 2015-12-09T03:42:37.060

3

(.*?) does not work for me. I am trying to match comments surrounded by /* */, which may contain multiple lines.

Try this:

([a]|[^a])

This regex matches a or anything else expect a. Absolutely, it means matching everything.

BTW, in my situation, /\*([a]|[^a])*/ matches C style comments.

Thank @mpen for a more concise way.

[\s\S]

edited Dec 09 '15 at 03:42

answered Sep 21 '15 at 12:47

oryxfea

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The most common way to do this in JS is `[\s\S]` -- i.e. match spaces and non-spaces. – mpen Dec 08 '15 at 23:41

score 3 · Answer 10 · edited Dec 13 '20 at 05:17

For JavaScript the best and simplest answer would seem to be /.\*/.

As suggested by others /(.*?)/ would work as well but /.\*/ is simpler. The () inside the pattern are not needed, as far as I can see nor the ending ? to match absolutely anything (including empty strings)

NON-SOLUTIONS:

/[\s\S]/ does NOT match empty strings so it's not the solution.
/[\s\S]\*/ DOES match also empty strings. But it has a problem: If you use it in your code then you can't comment out such code because the */ is interpreted as end-of-comment.

/([\s\S]\*)/ works and does not have the comment-problem. But it is longer and more complicated to understand than /.*/.

BTW. it seems the code-excerpts above make Stack Overflow render parts of my answer in italics, that was not my intention. — Panu Logic, Oct 24 '18 at 17:42

score 3 · Answer 11 · edited Jul 12 '20 at 22:05

The 2018 specification provides the s flag (alias: dotAll), so that . will match any character, including linebreaks:

const regExAll = /.*/s; //notice the 's'

let str = `
Everything
    in  this
            string
                    will
                        be
    matched. Including whitespace (even Linebreaks).
`;

console.log(`Match:`, regExAll.test(str)); //true
console.log(`Index Location:`, str.search(regExAll));

let newStr = str.replace(regExAll,"");
console.log(`Replaced with:`,newStr); //Index: 0

score 3 · Answer 12 · answered Jul 17 '20 at 06:20

Regex:
```
/I bought.*sheep./
```
Matches - the whole string till the end of line

I bought sheep. I bought a sheep. I bought five sheep.
Regex:
```
/I bought(.*)sheep./
```
Matches - the whole string and also capture the sub string within () for further use

I bought sheep. I bought a sheep. I bought five sheep.

I boughtsheep. I bought a sheep. I bought fivesheep.

Example using Javascript/Regex
```
'I bought sheep. I bought a sheep. I bought five sheep.'.match(/I bought(.*)sheep./)[0];
```
Output:

"I bought sheep. I bought a sheep. I bought five sheep."
```
'I bought sheep. I bought a sheep. I bought five sheep.'.match(/I bought(.*)sheep./)[1];
```
Output:

" sheep. I bought a sheep. I bought five "

score 0 · Answer 13 · answered May 29 '19 at 21:24

I recommend use /(?=.*...)/g

Example

const text1 = 'I am using regex';
/(?=.*regex)/g.test(text1) // true

const text2 = 'regex is awesome';
/(?=.*regex)/g.test(text2) // true

const text3 = 'regex is util';
/(?=.*util)(?=.*regex)/g.test(text3) // true

const text4 = 'util is necessary';
/(?=.*util)(?=.*regex)/g.test(text4) // false because need regex in text

Use regex101 to test

score 0 · Answer 14 · answered Jun 21 '19 at 01:23

0

Honestly alot of the answers are old so i found that if you simply just test any string regardless of character content with "/.*/i" will sufficiently get EVERYTHING.

answered Jun 21 '19 at 01:23

kjohnsonthecoder

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`/.*/i` will not match line breaks. Further, `/i` (the ‘ignore case’ flag) is redundant. – Danny Dec 16 '20 at 17:06

score 0 · Answer 15 · answered Jul 15 '11 at 19:11

0

<?php
$str = "I bought _ sheep";
preg_match("/I bought (.*?) sheep", $str, $match);
print_r($match);
?>

http://sandbox.phpcode.eu/g/b2243.php

answered Jul 15 '11 at 19:11

genesis

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Malcolm · Answer 16 · 2022-09-28T18:36:59.903

0

One option is the empty regular expression, denoted in JavaScript as /(?:)/. (You can also use new RegExp()). Logically, an empty regular expression should match strings containing "emptiness" at any position--which of course is all of them.

See this SO question and this blog post for discussion and more details.

edited Sep 28 '22 at 18:36

answered Sep 28 '22 at 18:31

Malcolm

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score -3 · Answer 17 · answered Jul 26 '21 at 22:54

-3

I use this: (.|\n)+ works like a charm for me!

answered Jul 26 '21 at 22:54

Montgomery Gole

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**Never** use this pattern (unless you have to in ElasticSearch regex flavor). It causes huge amount of backtrackings steps, and leads to stack overflow issues. Besides, this solution is mentioned [earlier](https://stackoverflow.com/a/56105382/3832970). – Wiktor Stribiżew Sep 06 '21 at 10:14

Regular Expressions- Match Anything

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