Swift: Adding leading zeroes certain numbers of an array [duplicate] - arrays

Question

I'd like to convert an Int in Swift to a String with leading zeros. For example consider this code:
for myInt in 1 ... 3 {
print("\(myInt)")
}
Currently the result of it is:
1
2
3
But I want it to be:
01
02
03
Is there a clean way of doing this within the Swift standard libraries?

Answer 1

Assuming you want a field length of 2 with leading zeros you'd do this:
import Foundation
for myInt in 1 ... 3 {
print(String(format: "%02d", myInt))
}
output:
01
02
03
This requires import Foundation so technically it is not a part of the Swift language but a capability provided by the Foundation framework. Note that both import UIKit and import Cocoa include Foundation so it isn't necessary to import it again if you've already imported Cocoa or UIKit.
The format string can specify the format of multiple items. For instance, if you are trying to format 3 hours, 15 minutes and 7 seconds into 03:15:07 you could do it like this:
let hours = 3
let minutes = 15
let seconds = 7
print(String(format: "%02d:%02d:%02d", hours, minutes, seconds))
output:
03:15:07

Answer 2

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")

Answer 3

For left padding add a string extension like this:
Swift 5.0 +
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"))
}

Answer 4

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"

Answer 5

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

Answer 6

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`

Answer 7

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

Answer 8

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"

Answer 9

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

Answer 10

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"

Answer 11

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)")
}

Answer 12

Unlike the other answers that use a formatter, you can also just add an "0" text in front of each number inside of the loop, like this:
for myInt in 1...3 {
println("0" + "\(myInt)")
}
But formatter is often better when you have to add suppose a designated amount of 0s for each seperate number. If you only need to add one 0, though, then it's really just your pick.