Using output of a regex to feed a function

Question

I'm trying to run a function taking as a parameter the matching criteria of a replace call, but I'm struggling with it. In the scenario of:

x="foo OFFSET_5 bar";
regex=/OFFSET_(-?[0-9]*)\b/g;
function timesThree(n){  return n*3 }

If I do a simple x.replace(regex, '$1') the output is, as expected, foo 5 bar. However, I want to pass that 5 to timesThree, I'm unable to, getting always NaNinstead of 15.

I tried x.replace(regex, timesThree($1)), but to no avail.

Where is my code failing?

Thanks in advance

@mplungjan - OP seems to want to use this as part of a replace. — T.J. Crowder
– T.J. Crowder, Commented Oct 16, 2019 at 9:43

Community · Accepted Answer · 2019-10-16 10:01:12Z

1

You'd do it with a wrapper function around the call:

x.replace(regex, (m, c0) => timesThree(c0));

Live Example:

const x = "foo OFFSET_5 bar";
const regex = /OFFSET_(-?[0-9]*)\b/g;
function timesThree(n){  return n*3 }
console.log(x.replace(regex, (m, c0) => timesThree(c0)));

or in ES5:

x.replace(regex, function(m, c0){ return timesThree(c0); });

Live Example:

var x = "foo OFFSET_5 bar";
var regex = /OFFSET_(-?[0-9]*)\b/g;
function timesThree(n){  return n*3 }
console.log(x.replace(regex, function(m, c0){ return timesThree(c0); }));

That works because the function is called with the overall match as the first argument (m, which we don't use), and then each capture group's captured text as a subsequent argument (c0 in this case). So we pass the first capture group's text into timesThree, which converts the string to number implicitly via *. The resulting number is implicitly converted back to string and inserted.

edited Oct 16, 2019 at 10:01

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answered Oct 16, 2019 at 9:35

T.J. Crowder

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1 Comment

T.J. Crowder Over a year ago

@ Anonymous User - Thanks for catching the const thing in the ES5!

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