Question

First of all, pardon my hacky code, I’m just trying to try this out and I’m learning javascript for the first time. Basically, Given the string “abc!random{3}” would mean return a string that starts with “abc” and ends with a random number from 0-3.

Here is what I have:

var pattern=/!random{([^{]*?)}/gi;
var text="abc!random{3}def!random{4}ghi!random{!random{3}}";

while (pattern.test(text))
{
    text=text.replace(pattern, Math.random() * parseInt("$1")); 
}

The problem is the parseInt function. It seems like the $1 does not get passed to it..the value of it gets cleared or something. If I do:

text=text.replace(pattern, "$1");   

It correctly returns what is in between the { }, so the regex is working and the match is being stored in $1. However, as soon as I use it as a parameter to $1, it seems like the value of it is cleared. What gives?

Answer 1

The second parameter to replace can either be a string to replace the entire match, within which, if present, all occurances of $1, $2, etc. are replaced with the captures OR it can be a function that takes two parameters (the match and the capture) and returns a replacement string for the capture:

var pattern=/!random{([^{]*?)}/gi;
var text="abc!random{3}def!random{4}ghi!random{!random{3}}";

text=text.replace(pattern,
    function(match,capture)
    {
        return Math.random() * parseInt(capture)
    });
WScript.echo(text);

Notice that the while loop is unnecessary: the regular expression already has the global flag set (“g”), which says to process the regex globally.

Also note that the function actually receives multiple (m) arguments: 1=the matched string, 2..m=the captures from left to right, m+2=the offset within the string where the match occured, and m+3=the entire string being matched. JavaScript allows the right-most arguments (of any function) to be omitted, which is why the example has only two.