我如何在Swift中生成一个随机的字母数字字符串?
当前回答
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)
}
其他回答
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。
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)
}
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
如果您只需要一个唯一标识符UUID()。uuidString可以满足您的需求。