Regular Expressions- Match Anything

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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.

16 Answers

/.*/ 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.

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

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

Hope it helps!

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.

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 /.*/.

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

  1. Regex:

    /I bought.*sheep./
    

    Matches - the whole string till the end of line

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

  2. 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 "

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

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

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