How can I split a string into an array of 2 character blocks in JS?

Question

For example, when given "hello world", I want [ "he", "ll", "o ", "wo", "rl", "d" ] returned.

I've tried "hello world".split(/../g) , but it didn't work, I just get empty strings:

 console.log("hello world".split(/../g));

I've also tried using a capture group which makes me closer to the goal:

 console.log("hello world".split(/(..)/g));

The only real solution I can think of here is to prune out the empty values with something like this:

 let arr = "hello world".split(/(..)/g); for (let i = 0; i < arr.length; i++) { if (arr[i] === "") arr.splice(i, 1); } console.log(arr);

Even though this works , it's a lot to read. Is there a possible way to do this in the split function?

Other attempted solutions

 let str = "hello world"; let arr = []; for (let i = 0; i < str.length; i++) { arr.push(str[i++] + (str[i] || '')); } console.log(arr);

Answer 1

Try this:

 const split = (str = "") => str.match(/.{1,2}/g) || []; console.log( split("hello world") );

Answer 2

You can use .match() to split string into two characters:

 var str = 'hello world'; console.log(str.match(/.{1,2}/g));

You can also increase number of spliting by editing {1,2} into number like 3 , 4 , 5 as in example:

3 characters split:

 var str = 'hello world'; console.log(str.match(/.{1,3}/g));

4 characters split:

 var str = 'hello world'; console.log(str.match(/.{1,4}/g));

Answer 3

It is easier to use a match, but if you must use split you can use your solution with a capture group to keep the text where you split on, and remove the empty entries afterwards.

The pattern will be (..?) to match 2 times any char where the second char is optional using the queation mark.

 console.log("hello world".split(/(..?)/).filter(Boolean));