Find and replace full words only in JavaScript using regex

Question

Say I had the following string:

var str = '(1 + foo + 3) / bar';

And I want to replace all strings just with the letter 'x' . I tried:

str = str.replace(/\w/g, 'x');

This results in:

(x + xxx + x) / xxx

Instead, I would like the result to be:

(1 + x + 3) / x

How would I do this? How would I find just the words that don't have digits and replace the word to a single letter?

Answer 1

Why not just use [az]+ instead of \\w ? (Make sure to add the case-insensetive flag, or use [a-zA-Z] instead)

Answer 2

str = str.replace(/\b[a-z]+\b/ig, 'x');

The \\b matches a word boundary. That way 'foo2' won't turn into 'x'. As others mentioned \\w includes numbers, but ALSO the underscore, so you won't want to use that. The i modifier does case insensitive matching (so that you can read a little easier).

Answer 3

采用：

str = str.replace(/[a-z]+/ig, 'x');

Answer 4

Try using [a-zA-Z] instead. \\w is equivalent to [a-zA-Z0-9_] .

str = str.replace(/[a-zA-Z]+/g, 'x');

Answer 5

You can try this regex:

str = str.replace(/[a-z]+/ig, 'x');

[az] - To indicate that you are looking for any letter.

+ To indicate that you are looking for a combination (xxx).

i To indicate that the text match can be case insensitive.

g - to indicate you are looking for all matches across the string.

or

you can use

   [a-zA-Z]

it will look for small letters az and capital letters AZ. This is for use without the case modifier.