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。

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

迅速:

let randomString = NSUUID().uuidString

为Swift 4更新。在类扩展上使用惰性存储变量。这只计算一次。

extension String {

    static var chars: [Character] = {
        return "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".map({$0})
    }()

    static func random(length: Int) -> String {
        var partial: [Character] = []

        for _ in 0..<length {
            let rand = Int(arc4random_uniform(UInt32(chars.count)))
            partial.append(chars[rand])
        }

        return String(partial)
    }
}

String.random(length: 10) //STQp9JQxoq

斯威夫特4

苹果推荐使用RandomNumberGenerator获得更好的性能

用法:String.random (20) 结果:CifkNZ9wy9jBOT0KJtV4

extension String{
   static func random(length:Int)->String{
        let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
        var randomString = ""

        while randomString.utf8.count < length{
            let randomLetter = letters.randomElement()
            randomString += randomLetter?.description ?? ""
        }
        return randomString
    }
}

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。