Regex matches in Javascript nesting issue

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?

James Hugard · Accepted Answer · 2009-05-23 16:22:01Z

5

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.

edited May 23, 2009 at 16:22

answered May 23, 2009 at 7:32

James Hugard

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3 Comments

BYK Over a year ago

Are you sure that the second parameter being a function is a standart implementation of JavaScript? If you can give a documentation link, that would be perfect since I have never came by a documentation likethis although it is a really cool feature =)

James Hugard Over a year ago

See section 15.5.4.11 of the Dec 1999 ECMAScript standard: ecma-international.org/publications/files/ECMA-ST/Ecma-262.pdf. Here is a direct link to the MSDN page describing replace, which documents the same thing: msdn.microsoft.com/en-us/library/t0kbytzc(VS.85).aspx.

Tate Johnson Over a year ago

stackoverflow.com/questions/61088/hidden-features-of-javascript/… helped me to understand why the function accepted multiple arguments in addition to James' answer.

Rudd Zwolinski · Accepted Answer · 2009-05-23 03:51:03Z

0

The replace method is operating on the value that it's passed. That is, if it gets a string containing $0, $1, etc., then it replaces that with the correct match. However, what you wrote is first parsing the string "$1" as an integer, multiplying it by a random number, and then passing that as the replacement. So it's almost certainly passing it NaN (due to being unable to parse "$1"), which is not what you want.

So, instead of

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

I would try

var matches = text.match(pattern);
text = text.replace(pattern, Math.random() * parseInt(matches[1]));

answered May 23, 2009 at 3:51

Rudd Zwolinski

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1 Comment

ryeguy Over a year ago

That didn't work. matches[1] just returns the entire string, not the captured group (which is what's in between the { })

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