Regular expressions - Replacing with XRegExp

Question

I have a feeling this is going to be a stupid question, but I'm completely baffled.

I'm using the XRegExp library for Javascript to generate a regular expression that matches any word in any alphabet, and does not match words with numbers, punctuation or special characters. The library is found here: http://xregexp.com/

This is what I have so far:

var tests = [
    "Hello",
    "Hello789",
    "Hello£*&£",
    "你好世界",
    "你好世界((£&"
];

var reg = XRegExp('^\\p{L}+$');

for (var i = 0; i < tests.length; i++) {
    console.log (tests[i] + " : " + (reg.test(tests[i]) ? "true" : "false"));
}

This produces the following output:

Hello : true
Hello789 : false
Hello£*&£ : false
你好世界 : true
你好世界((£& : false 

I know that reg is matching the right thing, but how do I use it in a replace so that the numbers, punctuation and special characters are stripped out? The output I want is :

Hello : Hello
Hello789 : Hello
Hello£*&£ : Hello
你好世界 : 你好世界
你好世界((£& : 你好世界

These are the things I've tried so far, with no success:

XRegExp.replace(tests[i], '^\\p{L}+$', ''));
XRegExp.replace(tests[i], '\\p{L}+$', ''));
XRegExp.replace(tests[i], '^\\p{L}', ''));
XRegExp.replace(tests[i], '\\p{L}', ''));

All of these return the same string I gave them, with no changes at all. Any one have an idea what I'm doing wrong?

Answer 1

Your regular expression is matching only letters, but it sounds like you want to match everything except letters:

var reg = XRegExp('\\P{L}+', 'g');

Then you can replace everything that matches with the empty string.

XRegExp.replace(tests[i], reg, '');