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。

这里有一个Swiftier语法的现成解决方案。你可以简单地复制粘贴它:

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
}

如果你喜欢一个框架，也有一些更方便的功能，然后随时签出我的项目handysswift。它还包括一个漂亮的随机字母数字字符串的解决方案:

String(randomWithLength: 8, allowedCharactersType: .alphaNumeric) // => "2TgM5sUG"

Swift 4.2更新

Swift 4.2在处理随机值和元素方面进行了重大改进。你可以在这里阅读更多关于这些改进的信息。以下是简化为几行的方法:

func randomString(length: Int) -> String {
  let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
  return String((0..<length).map{ _ in letters.randomElement()! })
}

Swift 3.0升级

func randomString(length: Int) -> String {

    let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    let len = UInt32(letters.length)

    var randomString = ""

    for _ in 0 ..< length {
        let rand = arc4random_uniform(len)
        var nextChar = letters.character(at: Int(rand))
        randomString += NSString(characters: &nextChar, length: 1) as String
    }

    return randomString
}

最初的回答:

func randomStringWithLength (len : Int) -> NSString {

    let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"

    var randomString : NSMutableString = NSMutableString(capacity: len)

    for (var i=0; i < len; i++){
        var length = UInt32 (letters.length)
        var rand = arc4random_uniform(length)
        randomString.appendFormat("%C", letters.characterAtIndex(Int(rand)))
    }

    return randomString
}

针对Swift 3.0

func randomString(_ length: Int) -> String {

    let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    let len = UInt32(letters.length)

    var randomString = ""

    for _ in 0 ..< length {
        let rand = arc4random_uniform(len)
        var nextChar = letters.character(at: Int(rand))
        randomString += NSString(characters: &nextChar, length: 1) as String
    }

    return randomString
}

来自任何字符集的纯Swift随机字符串。

用法:CharacterSet.alphanumerics。randomString(长度:100)

extension CharacterSet {
    /// extracting characters
    /// https://stackoverflow.com/a/52133647/1033581
    public func characters() -> [Character] {
        return codePoints().compactMap { UnicodeScalar($0) }.map { Character($0) }
    }
    public func codePoints() -> [Int] {
        var result: [Int] = []
        var plane = 0
        for (i, w) in bitmapRepresentation.enumerated() {
            let k = i % 8193
            if k == 8192 {
                plane = Int(w) << 13
                continue
            }
            let base = (plane + k) << 3
            for j in 0 ..< 8 where w & 1 << j != 0 {
                result.append(base + j)
            }
        }
        return result
    }

    /// building random string of desired length
    /// https://stackoverflow.com/a/42895178/1033581
    public func randomString(length: Int) -> String {
        let charArray = characters()
        let charArrayCount = UInt32(charArray.count)
        var randomString = ""
        for _ in 0 ..< length {
            randomString += String(charArray[Int(arc4random_uniform(charArrayCount))])
        }
        return randomString
    }
}

characters()函数是我所知道的最快的实现。

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。