我如何在Swift中生成一个随机的字母数字字符串?


当前回答

func randomUIDString(_ wlength: Int) -> String {

    let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    var randomString = ""

    for _ in 0 ..< wlength {
        let length = UInt32 (letters.length)
        let rand = arc4random_uniform(length)
        randomString = randomString.appendingFormat("%C", letters.character(at: Int(rand)));
    }

    return randomString
}

其他回答

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
    }
}

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

如果你的随机字符串应该是安全随机的,使用这个:

import Foundation
import Security

// ...

private static func createAlphaNumericRandomString(length: Int) -> String? {
    // create random numbers from 0 to 63
    // use random numbers as index for accessing characters from the symbols string
    // this limit is chosen because it is close to the number of possible symbols A-Z, a-z, 0-9
    // so the error rate for invalid indices is low
    let randomNumberModulo: UInt8 = 64

    // indices greater than the length of the symbols string are invalid
    // invalid indices are skipped
    let symbols = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"

    var alphaNumericRandomString = ""

    let maximumIndex = symbols.count - 1

    while alphaNumericRandomString.count != length {
        let bytesCount = 1
        var randomByte: UInt8 = 0

        guard errSecSuccess == SecRandomCopyBytes(kSecRandomDefault, bytesCount, &randomByte) else {
            return nil
        }

        let randomIndex = randomByte % randomNumberModulo

        // check if index exceeds symbols string length, then skip
        guard randomIndex <= maximumIndex else { continue }

        let symbolIndex = symbols.index(symbols.startIndex, offsetBy: Int(randomIndex))
        alphaNumericRandomString.append(symbols[symbolIndex])
    }

    return alphaNumericRandomString
}
func randomUIDString(_ wlength: Int) -> String {

    let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
    var randomString = ""

    for _ in 0 ..< wlength {
        let length = UInt32 (letters.length)
        let rand = arc4random_uniform(length)
        randomString = randomString.appendingFormat("%C", letters.character(at: 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