163

How can I do this one anywhere?

Basically, I am trying to match all kinds of miscellaneous characters such as ampersands, semicolons, dollar signs, etc.

Peter Mortensen
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Ryan Ward
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9 Answers9

297
[^a-zA-Z\d\s:]
  • \d - numeric class
  • \s - whitespace
  • a-zA-Z - matches all the letters
  • ^ - negates them all - so you get - non numeric chars, non spaces and non colons
Tudor Constantin
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42

This should do it:

[^a-zA-Z\d\s:]
Peter Mortensen
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Luke Sneeringer
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18

If you want to treat accented latin characters (eg. à Ñ) as normal letters (ie. avoid matching them too), you'll also need to include the appropriate Unicode range (\u00C0-\u00FF) in your regex, so it would look like this:

/[^a-zA-Z\d\s:\u00C0-\u00FF]/g
  • ^ negates what follows
  • a-zA-Z matches upper and lower case letters
  • \d matches digits
  • \s matches white space (if you only want to match spaces, replace this with a space)
  • : matches a colon
  • \u00C0-\u00FF matches the Unicode range for accented latin characters.

nb. Unicode range matching might not work for all regex engines, but the above certainly works in Javascript (as seen in this pen on Codepen).

nb2. If you're not bothered about matching underscores, you could replace a-zA-Z\d with \w, which matches letters, digits, and underscores.

Nick F
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  • This range contains some characters which are not alphanumeric (U+00D7 and U+00F7), and excludes a lot of valid accented characters from non-Western languages like Polish, Czech, Vietnamese etc. – tripleee Dec 06 '19 at 08:15
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    Upvoted for the description of each part of the RegEx. – morajabi Dec 09 '19 at 12:49
15

Try this:

[^a-zA-Z0-9 :]

JavaScript example:

"!@#$%* ABC def:123".replace(/[^a-zA-Z0-9 :]/g, ".")

See a online example:

http://jsfiddle.net/vhMy8/

Peter Mortensen
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Topera
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    Downvoted because this will not catch non-Latin characters, nor "special" Latin characters. – damian Jan 18 '16 at 08:19
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    It is easy to down vote an answer, and yet more difficult to provide constructive information to the board, e.g. how does one then catch non-Latin characters, nor "special" Latin characters? As of my count to here you have down voted 3 answers for the same reason, and in my opinion for a rather minor tweak. For example, I am here to find a regex for exactly what is discussed in these answers. I don't care about character sets that will not be used in my application. Law of diminishing returns. –  Jun 15 '16 at 13:16
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    Aaron might be a "minor tweak" to a US citizen, but highly relevant for... the rest of this planet. – Michael K. Borregaard Mar 17 '20 at 10:07
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    `[^a-zA-Z0-9 :]` can be replaced with `[^\w:]` – Moses Schwartz Aug 18 '20 at 17:40
5

If you mean "non-alphanumeric characters", try to use this:

var reg =/[^a-zA-Z0-9]/g      //[^abc]
Peter Mortensen
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Kim-Trinh
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4

No alphanumeric, white space or '_'. 

var reg = /[^\w\s)]|[_]/g;
Vasyl Gutnyk
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1

This regex works for C#, PCRE and Go to name a few.

It doesn't work for JavaScript on Chrome from what RegexBuddy says. But there's already an example for that here.

This main part of this is:

\p{L}

which represents \p{L} or \p{Letter} any kind of letter from any language.`

The full regex itself: [^\w\d\s:\p{L}]

Example: https://regex101.com/r/K59PrA/2

Peter Mortensen
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Ste
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  • This is the only answer here which deals correctly with Unicode accented alphabetics in a proper way. Sadly, not all regex engines support this facility (even Python lacks it, as of 3.8, even though its regex engine is ostensibly PCRE-based). – tripleee Dec 06 '19 at 08:30
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    I'll remove Python from the answer, I thought I tested that but apparently not. Thanks for pointing that out. – Ste Dec 06 '19 at 12:19
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In JavaScript:

/[^\w_]/g

^ negation, i.e. select anything not in the following set

\w any word character (i.e. any alphanumeric character, plus underscore)

_ negate the underscore, as it's considered a 'word' character

Usage example - const nonAlphaNumericChars = /[^\w_]/g;

Chris Halcrow
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0

Try to add this:

^[^a-zA-Z\d\s:]*$

This has worked for me... :)

Peter Mortensen
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Er Parthu
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  • This seems to repeat the accepted answer from 2011. The `^` and `$` anchors confines it to match entire lines and the `*` quantifier means it also matches empty lines. – tripleee Dec 06 '19 at 08:23