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
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
}
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)
}
斯威夫特5.0
// Generating Random String
func randomString(length: Int) -> String {
let letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
return String((0..<length).map{ _ in letters.randomElement()! })
}
// Calling to string
label.text = randomString(length: 3)
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("")
}
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"
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