Leading zeros for Int in Swift

With Swift 5, you may choose one of the three examples shown below in order to solve your problem.

#1. Using String's init(format:_:) initializer

Foundation provides Swift String a init(format:_:) initializer. init(format:_:) has the following declaration:

init(format: String, _ arguments: CVarArg...) 

Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted.

The following Playground code shows how to create a String formatted from Int with at least two integer digits by using init(format:_:):

import Foundation let string0 = String(format: "%02d", 0) // returns "00" let string1 = String(format: "%02d", 1) // returns "01" let string2 = String(format: "%02d", 10) // returns "10" let string3 = String(format: "%02d", 100) // returns "100" 

#2. Using String's init(format:arguments:) initializer

Foundation provides Swift String a init(format:arguments:) initializer. init(format:arguments:) has the following declaration:

init(format: String, arguments: [CVarArg]) 

Returns a String object initialized by using a given format string as a template into which the remaining argument values are substituted according to the user’s default locale.

The following Playground code shows how to create a String formatted from Int with at least two integer digits by using init(format:arguments:):

import Foundation let string0 = String(format: "%02d", arguments: [0]) // returns "00" let string1 = String(format: "%02d", arguments: [1]) // returns "01" let string2 = String(format: "%02d", arguments: [10]) // returns "10" let string3 = String(format: "%02d", arguments: [100]) // returns "100" 

#3. Using NumberFormatter

Foundation provides NumberFormatter. Apple states about it:

Instances of NSNumberFormatter format the textual representation of cells that contain NSNumber objects and convert textual representations of numeric values into NSNumber objects. The representation encompasses integers, floats, and doubles; floats and doubles can be formatted to a specified decimal position.

The following Playground code shows how to create a NumberFormatter that returns String? from a Int with at least two integer digits:

import Foundation let formatter = NumberFormatter() formatter.minimumIntegerDigits = 2 let optionalString0 = formatter.string(from: 0) // returns Optional("00") let optionalString1 = formatter.string(from: 1) // returns Optional("01") let optionalString2 = formatter.string(from: 10) // returns Optional("10") let optionalString3 = formatter.string(from: 100) // returns Optional("100") 

For left padding add a string extension like this:

extension String { func padLeft(totalWidth: Int, with byString: String) -> String { let toPad = totalWidth - self.count if toPad < 1 { return self } return "".padding(toLength: toPad, withPad: byString, startingAt: 0) + self } } 

Using this method:

for myInt in 1...3 { print("\(myInt)".padLeft(totalWidth: 2, with: "0")) } 

Swift 3.0+

Left padding String extension similar to padding(toLength:withPad:startingAt:) in Foundation

extension String { func leftPadding(toLength: Int, withPad: String = " ") -> String { guard toLength > self.characters.count else { return self } let padding = String(repeating: withPad, count: toLength - self.characters.count) return padding + self } } 

Usage:

let s = String(123) s.leftPadding(toLength: 8, withPad: "0") // "00000123" 

Using Swift 5’s fancy new extendible interpolation:

extension DefaultStringInterpolation { mutating func appendInterpolation(pad value: Int, toWidth width: Int, using paddingCharacter: Character = "0") { appendInterpolation(String(format: "%\(paddingCharacter)\(width)d", value)) } } let pieCount = 3 print("I ate \(pad: pieCount, toWidth: 3, using: "0") pies") // => `I ate 003 pies` print("I ate \(pad: 1205, toWidth: 3, using: "0") pies") // => `I ate 1205 pies` 

Swift 5

@imanuo answers is already great, but if you are working with an application full of number, you can consider an extension like this:

extension String { init(withInt int: Int, leadingZeros: Int = 2) { self.init(format: "%0\(leadingZeros)d", int) } func leadingZeros(_ zeros: Int) -> String { if let int = Int(self) { return String(withInt: int, leadingZeros: zeros) } print("Warning: \(self) is not an Int") return "" } } 

In this way you can call wherever:

String(withInt: 3) // prints 03 String(withInt: 23, leadingZeros: 4) // prints 0023 "42".leadingZeros(2) // prints 42 "54".leadingZeros(3) // prints 054 

The other answers are good if you are dealing only with numbers using the format string, but this is good when you may have strings that need to be padded (although admittedly a little diffent than the question asked, seems similar in spirit). Also, be careful if the string is longer than the pad.

 let str = "a str" let padAmount = max(10, str.count) String(repeatElement("-", count: padAmount - str.count)) + str 

Output "-----a str"

in Xcode 8.3.2, iOS 10.3 Thats is good to now

Sample1:

let dayMoveRaw = 5 let dayMove = String(format: "%02d", arguments: [dayMoveRaw]) print(dayMove) // 05 

Sample2:

let dayMoveRaw = 55 let dayMove = String(format: "%02d", arguments: [dayMoveRaw]) print(dayMove) // 55 

The below code generates a 3 digits string with 0 padding in front:

import Foundation var randomInt = Int.random(in: 0..<1000) var str = String(randomInt) var paddingZero = String(repeating: "0", count: 3 - str.count) print(str, str.count, paddingZero + str) 

Output:

5 1 005 88 2 088 647 3 647 

Swift 4* and above you can try this also:

func leftPadding(valueString: String, toLength: Int, withPad: String = " ") -> String { guard toLength > valueString.count else { return valueString } let padding = String(repeating: withPad, count: toLength - valueString.count) return padding + valueString } 

call the function:

leftPadding(valueString: "12", toLength: 5, withPad: "0") 

Output: "00012"

Details

Xcode 9.0.1, swift 4.0

Solutions

Data

import Foundation let array = [0,1,2,3,4,5,6,7,8] 

Solution 1

extension Int { func getString(prefix: Int) -> String { return "\(prefix)\(self)" } func getString(prefix: String) -> String { return "\(prefix)\(self)" } } for item in array { print(item.getString(prefix: 0)) } for item in array { print(item.getString(prefix: "0x")) } 

Solution 2

for item in array { print(String(repeatElement("0", count: 2)) + "\(item)") } 

Solution 3

extension String { func repeate(count: Int, string: String? = nil) -> String { if count > 1 { let repeatedString = string ?? self return repeatedString + repeate(count: count-1, string: repeatedString) } return self } } for item in array { print("0".repeate(count: 3) + "\(item)") }