How can I generate a random alphanumeric string in Swift?
If your random string should be secure-random, use this:
import Foundation
import Security
// ...
private static func createAlphaNumericRandomString(length: Int) -> String? {
// create random numbers from 0 to 63
// use random numbers as index for accessing characters from the symbols string
// this limit is chosen because it is close to the number of possible symbols A-Z, a-z, 0-9
// so the error rate for invalid indices is low
let randomNumberModulo: UInt8 = 64
// indices greater than the length of the symbols string are invalid
// invalid indices are skipped
let symbols = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"
var alphaNumericRandomString = ""
let maximumIndex = symbols.count - 1
while alphaNumericRandomString.count != length {
let bytesCount = 1
var randomByte: UInt8 = 0
guard errSecSuccess == SecRandomCopyBytes(kSecRandomDefault, bytesCount, &randomByte) else {
return nil
}
let randomIndex = randomByte % randomNumberModulo
// check if index exceeds symbols string length, then skip
guard randomIndex <= maximumIndex else { continue }
let symbolIndex = symbols.index(symbols.startIndex, offsetBy: Int(randomIndex))
alphaNumericRandomString.append(symbols[symbolIndex])
}
return alphaNumericRandomString
}
My even more Swift-ier implementation of the question:
func randomAlphanumericString(length: Int) -> String {
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".characters
let lettersLength = UInt32(letters.count)
let randomCharacters = (0..<length).map { i -> String in
let offset = Int(arc4random_uniform(lettersLength))
let c = letters[letters.startIndex.advancedBy(offset)]
return String(c)
}
return randomCharacters.joinWithSeparator("")
}
Simple and Fast -- UUID().uuidString
// Returns a string created from the UUID, such as "E621E1F8-C36C-495A-93FC-0C247A3E6E5F"
public var uuidString: String { get }
https://developer.apple.com/documentation/foundation/uuid
Swift 3.0
let randomString = UUID().uuidString //0548CD07-7E2B-412B-AD69-5B2364644433
print(randomString.replacingOccurrences(of: "-", with: ""))
//0548CD077E2B412BAD695B2364644433
EDIT
Please don't confuse with UIDevice.current.identifierForVendor?.uuidString
it won't give random values.
For folks who don't want to type out the entire set of characters:
func randomAlphanumericString(length: Int) -> String {
enum Statics {
static let scalars = [UnicodeScalar("a").value...UnicodeScalar("z").value,
UnicodeScalar("A").value...UnicodeScalar("Z").value,
UnicodeScalar("0").value...UnicodeScalar("9").value].joined()
static let characters = scalars.map { Character(UnicodeScalar($0)!) }
}
let result = (0..<length).map { _ in Statics.characters.randomElement()! }
return String(result)
}
With Swift 4.2 your best bet is to create a string with the characters you want and then use randomElement to pick each character:
let length = 32
let characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let randomCharacters = (0..<length).map{_ in characters.randomElement()!}
let randomString = String(randomCharacters)
I detail more about these changes here.
Here's a ready-to-use solution in Swiftier syntax. You can simply copy and paste it:
func randomAlphaNumericString(length: Int) -> String {
let allowedChars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let allowedCharsCount = UInt32(allowedChars.characters.count)
var randomString = ""
for _ in 0 ..< length {
let randomNum = Int(arc4random_uniform(allowedCharsCount))
let randomIndex = allowedChars.index(allowedChars.startIndex, offsetBy: randomNum)
let newCharacter = allowedChars[randomIndex]
randomString += String(newCharacter)
}
return randomString
}
If you prefer a Framework that also has some more handy features then feel free to checkout my project HandySwift. It also includes a beautiful solution for random alphanumeric strings:
String(randomWithLength: 8, allowedCharactersType: .alphaNumeric) // => "2TgM5sUG"