Remove all ANSI colors/styles from strings

Question

I use a library that adds ANSI colors / styles to strings. For example:

> \"Hello World\".rgb(255, 255, 255)
\'\\u001b[38;5;231mHello World\\u001b[0m\'
&g         


        

Answer 1

The regex you should be using is

/[\u001b\u009b][[()#;?]*(?:[0-9]{1,4}(?:;[0-9]{0,4})*)?[0-9A-ORZcf-nqry=><]/g

This matches most of the ANSI escape codes, beyond just colors, including the extended VT100 codes, archaic/proprietary printer codes, etc.

Note that the \u001b in the above regex may not work for your particular library (even though it should); check out my answer to a similar question regarding acceptable escape characters if it doesn't.

If you don't like regexes, you can always use the strip-ansi package.

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For instance, the string jumpUpAndRed below contains ANSI codes for jumping to the previous line, writing some red text, and then going back to the beginning of the next line - of which require suffixes other than m.

var jumpUpAndRed = "\x1b[F\x1b[31;1mHello, there!\x1b[m\x1b[E";
var justText = jumpUpAndRed.replace(
    /[\u001b\u009b][[()#;?]*(?:[0-9]{1,4}(?:;[0-9]{0,4})*)?[0-9A-ORZcf-nqry=><]/g, '');
console.log(justText);

Answer 2

The escape character is \u001b, and the sequence from [ until first m is encountered is the styling. You just need to remove that. So, replace globally using the following pattern:

/\u001b\[.*?m/g

Thus,

'\u001b[1m\u001b[38;5;231mHello World\u001b[0m\u001b[22m'.replace(/\u001b\[.*?m/g, '')