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
如果您只需要一个唯一标识符UUID()。uuidString可以满足您的需求。
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
}
来自任何字符集的纯Swift随机字符串。
用法:CharacterSet.alphanumerics。randomString(长度:100)
extension CharacterSet {
/// extracting characters
/// https://stackoverflow.com/a/52133647/1033581
public func characters() -> [Character] {
return codePoints().compactMap { UnicodeScalar($0) }.map { Character($0) }
}
public func codePoints() -> [Int] {
var result: [Int] = []
var plane = 0
for (i, w) in bitmapRepresentation.enumerated() {
let k = i % 8193
if k == 8192 {
plane = Int(w) << 13
continue
}
let base = (plane + k) << 3
for j in 0 ..< 8 where w & 1 << j != 0 {
result.append(base + j)
}
}
return result
}
/// building random string of desired length
/// https://stackoverflow.com/a/42895178/1033581
public func randomString(length: Int) -> String {
let charArray = characters()
let charArrayCount = UInt32(charArray.count)
var randomString = ""
for _ in 0 ..< length {
randomString += String(charArray[Int(arc4random_uniform(charArrayCount))])
}
return randomString
}
}
characters()函数是我所知道的最快的实现。
SwifterSwift有这个实现
/// SwifterSwift: Create a new random string of given length.
///
/// String(randomOfLength: 10) -> "gY8r3MHvlQ"
///
/// - Parameter length: number of characters in string.
init(randomOfLength length: Int) {
guard length > 0 else {
self.init()
return
}
let base = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
var randomString = ""
for _ in 1...length {
randomString.append(base.randomElement()!)
}
self = randomString
}
我做了一些改变,并使用这个实现
static func random(randomOfLength length: Int) -> String {
guard length > 0 else { return "" }
let base = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
var randomString = ""
for _ in 1...length {
guard let randomCharacter = base.randomElement() else { continue }
randomString.append(randomCharacter)
}
return randomString
}
为Swift 4更新。在类扩展上使用惰性存储变量。这只计算一次。
extension String {
static var chars: [Character] = {
return "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".map({$0})
}()
static func random(length: Int) -> String {
var partial: [Character] = []
for _ in 0..<length {
let rand = Int(arc4random_uniform(UInt32(chars.count)))
partial.append(chars[rand])
}
return String(partial)
}
}
String.random(length: 10) //STQp9JQxoq
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