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
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)
}
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
}
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。
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