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Regular Expressions- Match Anything

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.

.* should work. Can you paste your actual code?
What language are you coding in?
a dot won't match a newline
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.
(?s:.) - the inline modifier group match any char including line break chars. In your case, it would be something like this: (?s:.*?). Taken from the answer by Wiktor Stribiżew.

T
Tim Pietzcker

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.


.* doesn't match \n but it maches a string that contains only \n because it matches 0 character.
if using Javascript don't forget to slash the slash when setting a variable to this pattern eg: var pattern = "[\\s\\S]*";
the (?s).* works for me matched everything including new line
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.
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(".*")).
A
Adam Arold

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


but does that match a newline?
I used this on Sublime Text and it worked perfectly. Thanks! ("title":".*?")
In Javascript, use [\s\S]* or [^]*.
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.
This doesn't seem to include new lines or trailing full stops. I found this slight alteration hits everything (?s).*
T
Tyler

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


M
Mike Mozhaev

Try this:

I bought (.* )?sheep

or even

I bought .*sheep

S
Sam

/.*/ 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.


does it match no chars? Want match one or more.
Yes, the solutions I posted above will also match nothing/empty string. If you want to match at least one character, simply replace * with +.
/[\s\S]*/ works to me to match anything ignore break lines
S
Sang Huynh

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

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

Hope it helps!


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!)
Also, this will only work with LF (\n) line endings, but not CR (\r) or CRLF (\r\n) line endings.
J
Josh

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


S
Sam

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.


o
oryxfea

(.*?) 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]

The most common way to do this in JS is [\s\S] -- i.e. match spaces and non-spaces.
S
Sam

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.
S
Sam

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


S
SridharKritha

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 "


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

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


A
Alex Montoya

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


k
kjohnsonthecoder

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.


/.*/i will not match line breaks. Further, /i (the ‘ignore case’ flag) is redundant. 😉
M
Montgomery Gole

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


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.