Most efficient way to remove leading zeros from Swift 3 string

Question

I have a string such as "00123456" that I would like to have in an string "123456", with the leading zeros removed.

I've found several examples for Objective-C but not sure best way to do so with Swift 3.

Thanks

Good question! Moreover, to check this you have to testify with CPU time in async.main queue — J A S K I E R, Jul 19 '18 at 13:19

score 43 · Answer 1 · answered Aug 07 '18 at 10:24

43

You can do that with Regular Expression

let string = "00123456"
let trimmedString = string.replacingOccurrences(of: "^0+", with: "", options: .regularExpression)

The benefit is no double conversion and no force unwrapping.

answered Aug 07 '18 at 10:24

vadian

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This should be an excepted answer. – Roma Kavinskyi May 13 '22 at 03:56

score 18 · Accepted Answer · answered Nov 07 '17 at 17:50

18

Just convert the string to an int and then back to a string again. It will remove the leading zeros.

let numberString = "00123456"
let numberAsInt = Int(numberString)
let backToString = "\(numberAsInt!)"

Result: "123456"

answered Nov 07 '17 at 17:50

livtay

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Be careful here with the forced unwrap: make sure that the String is not empty ("") i.e. with a guard statement. Otherwise this code will crash. – Blackvenom Mar 06 '18 at 11:41
In this case the string is defined so it will never be nil. But yes, if the string is dynamic you would have to handle it to guard for a nil value. – livtay Mar 06 '18 at 14:01
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if the resulting number is too big this will fail. – jimijon Dec 17 '18 at 13:36
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why is this marked correct, it's obviously not the right type safe decision. – Schemetrical Jul 27 '19 at 00:47
Be careful with this implementation. I used it for a production code and it crushed because some user entered a number that exceeds the Int64 limit. `Int(numberString)` was nil in this case, and force-unwrapping caused a crash – Roma Kavinskyi May 13 '22 at 03:53

score 1 · Answer 3 · answered Jun 01 '18 at 12:16

First, create Validator then use it in any class. This is an example and it works :) This is swift 4.0

class PhoneNumberExcludeZeroValidator {
    func validate(_ value: String) -> String {
        var subscriberNumber = value
        let prefixCase = "0"
        if subscriberNumber.hasPrefix(prefixCase) {
            subscriberNumber.remove(at: subscriberNumber.startIndex)
        }
        return subscriberNumber
    }
}

example for usage:

if let countryCallingCode = countryCallingCodeTextField.text, var subscriberNumber = phoneNumberTextField.text {

     subscriberNumber = PhoneNumberExcludeZeroValidator().validate(subscriberNumber)

          let phoneNumber = "\(countryCallingCode)\(subscriberNumber)"
          registerUserWith(phoneNumber: phoneNumber)
}

This only removes first 0. If someone enters 0011234 then it will be like: 011234 — Neeraj Joshi, Dec 18 '19 at 13:23

score -1 · Answer 4 · answered Aug 07 '18 at 10:19

-1

let number = "\(String(describing: Int(text)!))"

answered Aug 07 '18 at 10:19

nikdange_me

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Why not simply `"\(Int(text)!)"` or `String(Int(text)!)`? Creating a string from a string with String Interpolation is redundant. – vadian Aug 20 '18 at 07:20

Most efficient way to remove leading zeros from Swift 3 string

4 Answers4