我如何在Swift中生成一个随机的字母数字字符串?


当前回答

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

其他回答

如果您只需要一个唯一标识符UUID()。uuidString可以满足您的需求。

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
}

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
    }
}

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

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
}