为了提供主题并显示快速下标的可能性，这里有一个基于“substring-toolbox”下标的小字符串

这些方法是安全的，并且不会遍历字符串索引

extension String {
    // string[i] -> one string char
    subscript(pos: Int) -> String { return String(Array(self)[min(self.length-1,max(0,pos))]) }

    // string[pos,len] -> substring from pos for len chars on the left
    subscript(pos: Int, len: Int) -> String { return self[pos, len, .pos_len, .left2right] }

    // string[pos, len, .right2left] -> substring from pos for len chars on the right
    subscript(pos: Int, len: Int, way: Way) -> String { return self[pos, len, .pos_len, way] }

    // string[range] -> substring form start pos on the left to end pos on the right
    subscript(range: Range<Int>) -> String { return self[range.startIndex, range.endIndex, .start_end, .left2right] }

    // string[range, .right2left] -> substring start pos on the right to end pos on the left
    subscript(range: Range<Int>, way: Way) -> String { return self[range.startIndex, range.endIndex, .start_end, way] }

    var length: Int { return countElements(self) }
    enum Mode { case pos_len, start_end }
    enum Way { case left2right, right2left }
    subscript(var val1: Int, var val2: Int, mode: Mode, way: Way) -> String {
        if mode == .start_end {
            if val1 > val2 { let val=val1 ; val1=val2 ; val2=val }
            val2 = val2-val1
        }
        if way == .left2right {
            val1 = min(self.length-1, max(0,val1))
            val2 = min(self.length-val1, max(1,val2))
        } else {
            let val1_ = val1
            val1 = min(self.length-1, max(0, self.length-val1_-val2 ))
            val2 = max(1, (self.length-1-val1_)-(val1-1) )
        }
        return self.bridgeToObjectiveC().substringWithRange(NSMakeRange(val1, val2))

        //-- Alternative code without bridge --
        //var range: Range<Int> = pos...(pos+len-1)
        //var start = advance(startIndex, range.startIndex)
        //var end = advance(startIndex, range.endIndex)
        //return self.substringWithRange(Range(start: start, end: end))
    }
}


println("0123456789"[3]) // return "3"

println("0123456789"[3,2]) // return "34"

println("0123456789"[3,2,.right2left]) // return "56"

println("0123456789"[5,10,.pos_len,.left2right]) // return "56789"

println("0123456789"[8,120,.pos_len,.right2left]) // return "01"

println("0123456789"[120,120,.pos_len,.left2right]) // return "9"

println("0123456789"[0...4]) // return "01234"

println("0123456789"[0..4]) // return "0123"

println("0123456789"[0...4,.right2left]) // return "56789"

println("0123456789"[4...0,.right2left]) // return "678" << because ??? range can wear endIndex at 0 ???

你也可以像这样将字符串转换为字符数组:

let text = "My Text"
let index = 2
let charSequence = text.unicodeScalars.map{ Character($0) }
let char = charSequence[index]

这是在常数时间内在指定索引处获取char的方法。

下面的示例不是在常数时间内运行，而是需要线性时间。所以如果你有很多搜索字符串索引使用上述方法。

let char = text[text.startIndex.advancedBy(index)]

斯威夫特4

String(Array(stringToIndex)[index]) 

这可能是一次性解决这个问题的最好方法。您可能希望首先将String转换为数组，然后再将结果转换为String。否则，将返回字符而不是字符串。

示例String(Array("HelloThere")[1])将返回"e"作为字符串。

(数组("HelloThere")[1]将返回"e"作为字符。

Swift不允许字符串像数组一样被索引，但这就完成了工作，用蛮力的方式。

在Swift 5中，不扩展字符串:

var str = "ABCDEFGH"
for char in str {
if(char == "C") { }
}

以上Swift代码与Java代码相同:

int n = 8;
var str = "ABCDEFGH"
for (int i=0; i<n; i++) {
if (str.charAt(i) == 'C') { }
}

在项目中包含此扩展

  extension String{
func trim() -> String
{
    return self.trimmingCharacters(in: NSCharacterSet.whitespaces)
}

var length: Int {
    return self.count
}

subscript (i: Int) -> String {
    return self[i ..< i + 1]
}

func substring(fromIndex: Int) -> String {
    return self[min(fromIndex, length) ..< length]
}

func substring(toIndex: Int) -> String {
    return self[0 ..< max(0, toIndex)]
}

subscript (r: Range<Int>) -> String {
    let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
                                        upper: min(length, max(0, r.upperBound))))
    let start = index(startIndex, offsetBy: range.lowerBound)
    let end = index(start, offsetBy: range.upperBound - range.lowerBound)
    return String(self[start ..< end])
}

func substring(fromIndex: Int, toIndex:Int)->String{
    let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
    let endIndex = self.index(startIndex, offsetBy: toIndex-fromIndex)

    return String(self[startIndex...endIndex])
}

然后像这样使用函数

let str = "Sample-String"

let substring = str.substring(fromIndex: 0, toIndex: 0) //returns S
let sampleSubstr = str.substring(fromIndex: 0, toIndex: 5) //returns Sample

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)