How to use an empty string in Regex Literal?

Question

I thought I'd space out after every letter in a string, for example foo becomes f o o

What I'm thinking is 'foo'.replace(//g, ' '); ( the g flag to replace every instance, otherwise I wouldn't want regex, would I :) ).

I only have one slight problem... in JavaScript, // is a comment, so it does not work.

How can I achieve this typing a regex literal (with slashes), or is it not possible and I would have to create (by typing) a Regex object?

@JonathanLonowski That would work, but /()/g is simpler because there's no capturing going on here. (Which is what I posted in my answer) — tckmn
– tckmn, Commented Jan 21, 2014 at 23:15
@DoorknobofSnow Sure. But, I wasn't mentioning it to compete. Just noting it as a common definition in various engines for an "empty" pattern. — Jonathan Lonowski
– Jonathan Lonowski, Commented Jan 21, 2014 at 23:32

Stephan · Accepted Answer · 2014-01-21 23:28:55Z

2

Try this regex instead:

/.(?!$)/g

You can read it as: "Find all char except the last one."

Description

Regular expression visualization

Sample code

'foo'.replace(/.(?!$)/g, '$& ');

Demo

http://regex101.com/r/xH9bK1

edited Jan 21, 2014 at 23:28

answered Jan 21, 2014 at 23:16

Stephan

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tckmn · Accepted Answer · 2014-01-21 23:10:24Z

1

There are a few ways you could do this. (These methods are ranked from best practice to worst. I recommend the first one, and I highly discourage the last.)

Use split/join, which would be a better method with less overhead. No need for a regex here:

'foo'.split('').join(' ');

Or use the RegExp constructor

'foo'.replace(new RegExp('', 'g'), ' ');

Or, add a useless group (not recommended because it's unclear at first glance):

'foo'.replace(/()/g, ' ');

One more comment, based on your last paragraph:

How can I achieve this using a regex literal, or is it not possible and I would have to create a Regex object?

They are literally the same thing:

screenshot

"using a regex literal... or... Regex object" doesn't make sense, because a regex literal is a RegExp object.

edited Jan 21, 2014 at 23:10

answered Jan 21, 2014 at 23:04

tckmn

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4 Comments

elclanrs Over a year ago

The regex might be faster in some browsers, here's a useless test jsperf.com/split-join-vs-regex-literal-replace

Stephan Over a year ago

The useless group approach will add a space before and after the string.See here: regex101.com/r/pT9mT5

tckmn Over a year ago

@Alex It's never mentioned that's unwanted... in fact, that's what would happen if you could use // as a regex literal.

Stephan Over a year ago

@DoorknobofSnow OP states on the very first line (at the end): for example foo becomes f o o

Gabriele Petrioli · Accepted Answer · 2014-01-21 23:21:33Z

0

You could use the \B (everything but word boundary)

'foo'.replace(/\B/g,' ');

or if you want to replace at every place

'foo'.replace(/\B|\b/g,' ');

edited Jan 21, 2014 at 23:21

answered Jan 21, 2014 at 23:13

Gabriele Petrioli

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1 Comment

tckmn Over a year ago

this will change, for example, foo bar to f o o b a r (instead of the correct result, f o o [space] b a r). Also it's even more unclear and hard to understand. :P

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