我如何在Swift中生成一个随机的字母数字字符串?
当前回答
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
}
其他回答
Swift 4.2更新
Swift 4.2在处理随机值和元素方面进行了重大改进。你可以在这里阅读更多关于这些改进的信息。以下是简化为几行的方法:
func randomString(length: Int) -> String {
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
return String((0..<length).map{ _ in letters.randomElement()! })
}
Swift 3.0升级
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
}
最初的回答:
func randomStringWithLength (len : Int) -> NSString {
let letters : NSString = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
var randomString : NSMutableString = NSMutableString(capacity: len)
for (var i=0; i < len; i++){
var length = UInt32 (letters.length)
var rand = arc4random_uniform(length)
randomString.appendFormat("%C", letters.characterAtIndex(Int(rand)))
}
return randomString
}
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"
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
迅速:
let randomString = NSUUID().uuidString
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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