replace string with numbers that resembles the string using string.replace()

Question

Here I have a line of code which I would prefer to take '123' as a string value.
Then with the help of split() and join() it will become a string like '1,2,3'.
Then I want to replace all the numeric string values with actual number like 1,2.....
For this I used a variable called k and give it an initial value 1.
I used an increment operator inside replace to change the value of k.
But it is not working. Maybe I am getting it wrong how string.replace() execute the replacement operation.
I also tried to use parseInt('$1') it also didn't work.

console gives output of 1,1,1.No increment operation takes place.

<html>
<body>
<script>

(function(){
var k=1;
   console.log(prompt('input here').split('').join(',').replace(/(\d)/g,k++));
})();
</script>
</body>
</html>

You mean you want to convert "123" into an array of numbers [1, 2, 3]? — Andy
– Andy, Commented Sep 12, 2014 at 12:48
If you don't want to increment the numbers var arr = '123'.split('').map(Number); is the simplest solution I know. — Andy
– Andy, Commented Sep 12, 2014 at 12:56
var a = "123".split('').map(function(item) { item = parseInt(item, 10); return ++item; }); works for me — mplungjan
– mplungjan, Commented Sep 12, 2014 at 12:57
Or, upon re-reading, is it actually and OP doesn't want to create incrementing numbers? — Bergi
– Bergi, Commented Sep 12, 2014 at 12:59

raina77ow · Accepted Answer · 2014-09-12 13:04:39Z

3

Well, you can use replace callback instead:

var k = 1;
'123'.split('').join(',').replace(/\d/g, function() {
  return k++;
});

As it stands, k++ value is taken as an argument - in other words, this operation is evaluated just once.

But if you want just to get a sequence of integers, there are far better approaches. For example:

function getSequence(start, end) {
  var res = [];
  for (var i = start; i <= end; i++) {
    res.push(i);
  }
  return res;

  // or just...
  // return Array.apply(0, Array(end - start + 1))
  //             .map(function(_, i) { return start + i });
  // ... if you feel adventurous. )
}
getSequence(2, 4); // [2, 3, 4]

edited Sep 12, 2014 at 13:04

answered Sep 12, 2014 at 12:52

raina77ow

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9 Comments

Bergi Over a year ago

You're relieving me from having to write my own answers :-)

JLRishe Over a year ago

Won't this only work for a string of the form "1234..."? I think the one in OP's post was just an example.

raina77ow Over a year ago

@JLRishe No, it replaces any digit in the number by the iterated index (leaving the non-digit symbols as they are).

raina77ow Over a year ago

Because k++ is just an expression, not a command. replace function will receive the result of this expression as its parameter - it won't have any idea about how you've come with this result.

raina77ow Over a year ago

The callback will be invoked each time a replacement operation is in progress; its return value for the given invokation will be used as a replacement value. That's why it's named callback, after all.

|

Andy · Accepted Answer · 2014-09-12 12:58:02Z

2

If you don't want to increase the numbers:

var arr = '123'.split('').map(Number);

If you do:

var arr = '123'.split('').map(function (num) { return Number(num) + 1; });

answered Sep 12, 2014 at 12:58

Andy

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