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

Swift 2.2版本

// based on https://gist.github.com/samuel-mellert/20b3c99dec168255a046
// which is based on https://gist.github.com/szhernovoy/276e69eb90a0de84dd90
// Updated to work on Swift 2.2

func randomString(length: Int) -> String {
    let charactersString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    let charactersArray : [Character] = Array(charactersString.characters)

    var string = ""
    for _ in 0..<length {
        string.append(charactersArray[Int(arc4random()) % charactersArray.count])
    }

    return string
}

基本上调用这个方法，它将生成一个随机字符串，长度为传递给函数的整数。要更改可能的字符，只需编辑charactersString字符串。也支持unicode字符。

https://gist.github.com/gingofthesouth/54bea667b28a815b2fe33a4da986e327

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
}

免费循环，尽管它被限制在43个字符。如果你需要更多，可以修改。与单独使用UUID相比，这种方法有两个优点:

“更大的熵”使用小写字母，因为UUID()只生成大写字母 UUID最大长度为36个字符(包括4个连字符)，不包含连字符的长度为32个字符。你应该需要更长的东西，或不希望连字符包括，使用base64EncodedString处理这个

此外，该函数使用UInt来避免负数。

 func generateRandom(size: UInt) -> String {
        let prefixSize = Int(min(size, 43))
        let uuidString = UUID().uuidString.replacingOccurrences(of: "-", with: "")
        return String(Data(uuidString.utf8)
            .base64EncodedString()
            .replacingOccurrences(of: "=", with: "")
            .prefix(prefixSize))
    }

在循环中调用它来检查输出:

for _ in 0...10 {
    print(generateRandom(size: 32))
}

生产:

Nzk3NjgzMTdBQ0FBNDFCNzk2MDRENzZF
MUI5RURDQzE1RTdCNDA3RDg2MTI4QkQx
M0I3MjJBRjVFRTYyNDFCNkI5OUM1RUVC
RDA1RDZGQ0IzQjI1NDdGREI3NDgxM0Mx
NjcyNUQyOThCNzhCNEVFQTk1RTQ3NTIy
MDkwRTQ0RjFENUFGNEFDOTgyQTUxODI0
RDU2OTNBOUJGMDE4NDhEODlCNEQ1NjZG
RjM2MTUxRjM4RkY3NDU2OUFDOTI0Nzkz
QzUwOTE1N0U1RDVENDE4OEE5NTM2Rjcy
Nzk4QkMxNUJEMjYwNDJDQjhBQkY5QkY5
ODhFNjU0MDVEMUI2NEI5QUIyNjNCNkVF

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
    } 
}