我如何在Swift中生成一个随机的字母数字字符串?


当前回答

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

其他回答

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
    } 
}

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

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)
}

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("")
}

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。