我如何在Swift中生成一个随机的字母数字字符串?


当前回答

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)
}

其他回答

迅速:

let randomString = NSUUID().uuidString

在Swift 4.2中,你最好的方法是创建一个包含你想要的字符的字符串,然后使用randomElement来选择每个字符:

let length = 32
let characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let randomCharacters = (0..<length).map{_ in characters.randomElement()!}
let randomString = String(randomCharacters)

我将在这里详细介绍这些变化。

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)
}

为Swift 4更新。在类扩展上使用惰性存储变量。这只计算一次。

extension String {

    static var chars: [Character] = {
        return "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".map({$0})
    }()

    static func random(length: Int) -> String {
        var partial: [Character] = []

        for _ in 0..<length {
            let rand = Int(arc4random_uniform(UInt32(chars.count)))
            partial.append(chars[rand])
        }

        return String(partial)
    }
}

String.random(length: 10) //STQp9JQxoq

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