顺便说一句，有几个函数可以直接应用于String的字符链表示，像这样:

var string = "Hello, playground"
let firstCharacter = string.characters.first // returns "H"
let lastCharacter = string.characters.last // returns "d"

结果类型为Character，但可以将其转换为String。

或:

let reversedString = String(string.characters.reverse())
// returns "dnuorgyalp ,olleH" 

:-)

对我来说最有效的方法是:

var firstName = "Olivia"
var lastName = "Pope"

var nameInitials.text = "\(firstName.prefix(1))" + "\    (lastName.prefix(1))"

输出:“OP”

使用字符就可以了。您可以快速地将字符串转换为字符数组，可以由CharacterView方法操作。

例子:

let myString = "Hello World!"
let myChars  = myString.characters

(完整的CharacterView文档)

(在Swift 3中测试)

不使用整数进行索引，只使用String.Index。主要是线性复杂度。您还可以从String中创建范围。索引并使用它们获取子字符串。

斯威夫特3.0

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.index(before: someString.endIndex)]
let charAtIndex = someString[someString.index(someString.startIndex, offsetBy: 10)]

let range = someString.startIndex..<someString.index(someString.startIndex, offsetBy: 10)
let substring = someString[range]

快2.倍

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.endIndex.predecessor()]
let charAtIndex = someString[someString.startIndex.advanceBy(10)]

let range = someString.startIndex..<someString.startIndex.advanceBy(10)
let subtring = someString[range]

请注意，不能使用从一个字符串到另一个字符串创建的索引(或范围)

let index10 = someString.startIndex.advanceBy(10)

//will compile
//sometimes it will work but sometimes it will crash or result in undefined behaviour
let charFromAnotherString = anotherString[index10]

斯威夫特5.1

这可能是所有答案中最简单的一个。

添加这个扩展:

extension String {
    func retrieveFirstCharacter() -> String? {
        guard self.count > 0 else { return nil }
        let numberOfCharacters = self.count
        return String(self.dropLast(numberOfCharacters - 1))
    }
}

你可以通过将字符串转换为数组，并通过下标获得特定的索引，如下所示

var str = "Hello"
let s = Array(str)[2]
print(s)