迅速:

let randomString = NSUUID().uuidString

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
}

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"

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
    } 
}

SwifterSwift有这个实现

/// SwifterSwift: Create a new random string of given length.
///
///     String(randomOfLength: 10) -> "gY8r3MHvlQ"
///
/// - Parameter length: number of characters in string.
init(randomOfLength length: Int) {
    guard length > 0 else {
        self.init()
        return
    }

    let base = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    var randomString = ""
    for _ in 1...length {
        randomString.append(base.randomElement()!)
    }
    self = randomString
}

我做了一些改变，并使用这个实现

static func random(randomOfLength length: Int) -> String {
    guard length > 0 else { return "" }
    let base = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    var randomString = ""
    for _ in 1...length {
        guard let randomCharacter = base.randomElement() else { continue }
        randomString.append(randomCharacter)
    }
    return randomString
}

斯威夫特5.6

此函数生成一个以36为基数的10位数字，然后将其作为字母数字字符串返回。

func randomCode(length: Int) -> String {
    let radix = 36 // = 10 digits + 26 letters
    let number = Int.random(in: 0..<(pow(radix, length)))
    return String(number, radix: radix, uppercase: true)
}

或者如果你不希望代码以“0”开头:

func randomCode(length: Int) -> String {
    let radix = 36 // = 10 digits + 26 letters
    let range = (pow(radix, length)/2)..<(pow(radix, length))
    let number = Int.random(in: range)
    return String(number, radix: radix, uppercase: true)
}