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)

更新后的2019年。

在不寻常的情况下

性能很重要。

下面是一个非常清晰的缓存函数:

func randomNameString(length: Int = 7)->String{
    
    enum s {
        static let c = Array("abcdefghjklmnpqrstuvwxyz12345789")
        static let k = UInt32(c.count)
    }
    
    var result = [Character](repeating: "-", count: length)
    
    for i in 0..<length {
        let r = Int(arc4random_uniform(s.k))
        result[i] = s.c[r]
    }
    
    return String(result)
}

这适用于当您有一个固定的、已知的字符集时。

方便的提示:

注意，“abcdefghjklmnpqrstuvwxyz12345789”避免了“坏”字符

没有0,o, o, i等等…人类经常混淆的字符。

这通常用于预订代码和人类客户将使用的类似代码。

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
    }
}

如果你的随机字符串应该是安全随机的，使用这个:

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
}

回答“我需要随机字符串”问题的问题(无论哪种语言)实际上每个解决方案都使用了有缺陷的字符串长度的主要规范。问题本身很少揭示为什么需要随机字符串，但我想挑战你很少需要长度为8的随机字符串。您总是需要一些唯一的字符串，例如，用于某种目的的标识符。

有两种主要的方法来获得严格唯一的字符串:确定性(不是随机的)和存储/比较(繁琐的)。我们该怎么办?我们放弃了。我们用概率唯一性来代替。也就是说，我们接受存在一些(无论多么小)的风险，即我们的字符串不是唯一的。这就是为什么理解碰撞概率和熵是有帮助的。

So I'll rephrase the invariable need as needing some number of strings with a small risk of repeat. As a concrete example, let's say you want to generate a potential of 5 million IDs. You don't want to store and compare each new string, and you want them to be random, so you accept some risk of repeat. As example, let's say a risk of less than 1 in a trillion chance of repeat. So what length of string do you need? Well, that question is underspecified as it depends on the characters used. But more importantly, it's misguided. What you need is a specification of the entropy of the strings, not their length. Entropy can be directly related to the probability of a repeat in some number of strings. String length can't.

这就是像EntropyString这样的库可以帮助的地方。使用EntropyString在500万个字符串中生成重复概率小于1万亿分之一的随机id:

import EntropyString

let random = Random()
let bits = Entropy.bits(for: 5.0e6, risk: 1.0e12)
random.string(bits: bits)

" Rrrj6pN4d6GBrFLH4 "

EntropyString默认使用32个字符的字符集。还有其他预定义的字符集，您也可以指定自己的字符。例如，生成与上面相同熵的id，但使用十六进制字符:

import EntropyString

let random = Random(.charSet16)
let bits = Entropy.bits(for: 5.0e6, risk: 1.0e12)
random.string(bits: bits)

“135fe71AEC7A80C02DCE5”

请注意由于所使用的字符集中字符总数的不同而导致的字符串长度的差异。在指定数量的潜在字符串中重复的风险相同。字符串长度不是。最重要的是，重复的风险和字符串的潜在数量是明确的。没有更多的猜测字符串长度。