在Swift 4.2中，你最好的方法是创建一个包含你想要的字符的字符串，然后使用randomElement来选择每个字符:

let length = 32
let characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let randomCharacters = (0..<length).map{_ in characters.randomElement()!}
let randomString = String(randomCharacters)

我将在这里详细介绍这些变化。

斯威夫特5.0

// Generating Random String
func randomString(length: Int) -> String {
    let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    return String((0..<length).map{ _ in letters.randomElement()! })
}
// Calling to string
label.text = randomString(length: 3)

如果您只需要一个唯一标识符UUID()。uuidString可以满足您的需求。

简单快捷——UUID().uuidString

//返回由UUID创建的字符串，例如"E621E1F8-C36C-495A-93FC-0C247A3E6E5F" uuidString:字符串{get} https://developer.apple.com/documentation/foundation/uuid

斯威夫特3.0

let randomString = UUID().uuidString //0548CD07-7E2B-412B-AD69-5B2364644433
print(randomString.replacingOccurrences(of: "-", with: ""))
//0548CD077E2B412BAD695B2364644433

EDIT

请不要与UIDevice.current.identifierForVendor混淆。uuidString它不会给出随机值。

一种避免输入整套字符的方法:

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)
}

迅速:

let randomString = NSUUID().uuidString