Format and replace a string with a regular expression

Question

I have a number that's at least 7 digits long. Typical examples: 0000123, 00001234, 000012345

I want to transform them so that they become respectively: 01:23, 12:34, 23:45

Which mean replacing the whole string by the last 4 characters and putting a colon in the middle.

I can get the last 4 digits with (\d{4})$ And I can get 2 groups with this: (\d{2})(\d{2})$

With the last option, on a string 0000123 $1:$2 match gives me 00001:23 where I want 01:23

I replace the string like so:

newVal = val.replace(/regex/, '$1:$2');

No need for an RE: time = val.substr(val.length - 4, 2) + ":" + val.substr(-2); ? — Alex K.
– Alex K., Commented Jul 28, 2017 at 13:03
@AlexK.: Since you didn't post an answer, I posted a CW with that, as it frankly seems the simplest, most straight-forward way to do it. But if you decide to post it, let me know and I'll delete it. — T.J. Crowder
– T.J. Crowder, Commented Jul 28, 2017 at 13:11

Wiktor Stribiżew · Accepted Answer · 2017-07-28 13:03:10Z

2

You need to match the beginning digits with \d* (or with just .* if there can be anything):

var val = "0001235";
var newVal = val.replace(/^\d*(\d{2})(\d{2})$/, '$1:$2');
console.log(newVal);

Pattern details:

^ - start of string
\d* - 0+ digits (or .* will match any 0+ chars other than line break chars)
(\d{2}) - Group 1 capturing 2 digits
(\d{2}) - Group 2 capturing 2 digits
$ - end of string.

answered Jul 28, 2017 at 13:03

Wiktor Stribiżew

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2 Comments

Duncan Over a year ago

Or as the digit string is known to be at least 7 digits you could match it with \d{3,}(\d{2})(\d{2}) and then any shorter digit strings will be skipped.

Wiktor Stribiżew Over a year ago

Yes, that is true.

T.J. Crowder · Accepted Answer · 2017-07-28 13:09:49Z

2

As Alex K. said, no need for a regular expression, just extract the parts you need with substr:

val = val.substr(-4, 2) + ":" + val.substr(-2);

Note that when the starting index is negative, it's from the end of the string.

Example:

function update(val) {
  return val.substr(-4, 2) + ":" + val.substr(-2);
}
function test(val) {
  console.log(val + " => " + update(val));
}
test("0000123");
test("0001234");
test("000012345");

answered Jul 28, 2017 at 13:09

community wiki

T.J. Crowder

3 Comments

gyc Over a year ago

Correct, this is a good solution. In my particular case there are other regex manipulations above my code and I feel the regex version will be helpful if the parsing becomes more complex later in the project.

T.J. Crowder Over a year ago

@gyc: Fair enough, but frankly, I'd introduce it later in the project if it becomes necessary, and keep it simple until then.

gyc Over a year ago

Yes definitely. Thanks

Nina Scholz · Accepted Answer · 2017-07-28 13:04:39Z

1

You could throw the first characters away and the replace only the last matched parts.

console.log('00000001234'.replace(/^(.*)(\d{2})(\d{2})$/, '$2:$3'));

answered Jul 28, 2017 at 13:04

Nina Scholz

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Comments

Daria Pydorenko · Accepted Answer · 2017-07-28 13:12:28Z

1

Use this regex: ^(\d+?)(\d{2})(\d{2})$:

var newVal = "0000123".replace(/^(\d+?)(\d{2})(\d{2})$/, '$2:$3');
console.log(newVal);

edited Jul 28, 2017 at 13:12

answered Jul 28, 2017 at 13:04

Daria Pydorenko

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