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
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)
}
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
}
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
}
如果您只需要一个唯一标识符UUID()。uuidString可以满足您的需求。