我如何在Swift中生成一个随机的字母数字字符串?
当前回答
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
}
}
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
}
}
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("")
}
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)
}
这里有一个Swiftier语法的现成解决方案。你可以简单地复制粘贴它:
func randomAlphaNumericString(length: Int) -> String {
let allowedChars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
let allowedCharsCount = UInt32(allowedChars.characters.count)
var randomString = ""
for _ in 0 ..< length {
let randomNum = Int(arc4random_uniform(allowedCharsCount))
let randomIndex = allowedChars.index(allowedChars.startIndex, offsetBy: randomNum)
let newCharacter = allowedChars[randomIndex]
randomString += String(newCharacter)
}
return randomString
}
如果你喜欢一个框架,也有一些更方便的功能,然后随时签出我的项目handysswift。它还包括一个漂亮的随机字母数字字符串的解决方案:
String(randomWithLength: 8, allowedCharactersType: .alphaNumeric) // => "2TgM5sUG"
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。
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