Can the \w regular expression pattern be modified to include special characters like é? If not, what other options are available?

Imagine having a regular expression that appears as follows:

\w+

In this case, the string "helloworld" would be accepted:

helloworld

However, "héllowörld" would not pass the test:

héllowörld

The regex will stop at é (and also break atö) even though to a human, héllowörld doesn't seem too far-fetched as a complete word.

Is there a way to enhance the functionality of \w to include special word characters? Or must every unique Latin character be added manually to the regex pattern like so:

[\wéèåöä...........]+

The idea of finding and adding all possible special Latin characters does not appear to be the most efficient solution.

What other alternatives are available?

Answer №1

\w will capture any word character within the [a-zA-Z0-9_] range, excluding non-English characters.

To find a regular expression that matches non-English characters, refer to this post: Regular expression to match non-English characters?

Answer №2

At times, I employ an inverse technique to identify non-English characters mixed in with others. Take a look

let string = "你好 κόσμος привет šđčߣłćž çë asgfgrtzj 657 #$%&/()=?*!";

This is the pattern used

let pattern = /([^0-9]+)/gi;

which will eliminate all numbers

你好 κόσμος привет šđčߣłćž çë asgfgrtzj #$%&/()=?*!";

by incorporating special characters from above into the pattern

let pattern = /([^0-9#$%&/()=?*!]+)/gi;

resulting in the final output shown below

你好 κόσμος привет šđčߣłćž çë asgfgrtzj 

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