这是我能想到的最快的解决办法。斯威夫特3.0

extension String {
    static func random(length: Int) -> String {
        let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
        let randomLength = UInt32(letters.characters.count)

        let randomString: String = (0 ..< length).reduce(String()) { accum, _ in
            let randomOffset = arc4random_uniform(randomLength)
            let randomIndex = letters.index(letters.startIndex, offsetBy: Int(randomOffset))
            return accum.appending(String(letters[randomIndex]))
        }

        return randomString
    } 
}

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"

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
}

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。

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
}

我对这个问题的更快速的回答是:

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("")
}