Regex replace in Javascript

Question

I have the following elements:

var url = "http://my.url/$1/$2/$3";
var params = ["unused", "first", "second", "third"];

I want to replace each $n element in the URL for the element in position n in the params array (i.e. $1 will be "first", $2 will be "second" and $3 will be "third").

I have the following code:

var cont=1;
while (params[cont]) {
    url = url.replace(new RegExp("\\$" + cont, "gm"), params[cont])
    cont++
}

The above code works, but I wonder if there would be a better way to do this replace (without looping).

Thank you very much in advance

Yury Tarabanko · Accepted Answer · 2014-08-20 10:23:07Z

4

Sure there is

url = url.replace(/\$(\d)/g, function(_, i) {
   return params[i];
});

UPD (pedantic mode): As @undefined pointed out there is still a loop. But an implicit one implemented by String.prototype.replace function for regexps having g flag on.

edited Aug 20, 2014 at 10:23

answered Aug 20, 2014 at 10:03

Yury Tarabanko

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4 Comments

Ram Over a year ago

Strictly speaking this is a loop as well.

Yury Tarabanko Over a year ago

@undefined Atleast not an explicit one. I doubt one can do multiple replaces (of unknow number of tokens) without some sort of a loop.

Ram Over a year ago

Yes, and that's the answer of the question.

Genzotto Over a year ago

Your code worked and even using an implicit loop, I like it more than mine because of its simplicity. Thank you

Jongware · Accepted Answer · 2014-08-20 10:03:43Z

2

Use a replace function:

var url = "http://my.url/$1/$2/$3";
var params = [ "first", "second", "third"];

url = url.replace (/\$(\d)/g, function(a,b) { return params[Number(b)-1]; });

As you can see, there is no need to use a "unused" element.

answered Aug 20, 2014 at 10:03

Jongware

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3 Comments

Genzotto Over a year ago

But this "unused" element is comming in my params array. It's an element that doesn't need to be replace but it's used to do an extra processing later, so it must be skipped in the replace process.

Jongware Over a year ago

In that case you can use Yury's code. Without my adjustment it's totally similar.

Genzotto Over a year ago

Yours is good too (basically it's the same), but unfortunately two answers cannot be accepted... However, there's my upvote.

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