我如何在Swift中生成一个随机的字母数字字符串?
当前回答
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
}
}
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。
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
}
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。
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
}
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
}
}
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