如果您正在寻找简单的方法来获得字符或字符串的索引，请检查这个库http://www.dollarswift.org/#indexof-char-character-int

您也可以使用另一个字符串或正则表达式模式从字符串中获取indexOf

String是NSString的桥接类型，add

import Cocoa

到你的swift文件，并使用所有“旧”的方法。

你可以用这个找到字符串中一个字符的索引号:

var str = "abcdefghi"
if let index = str.firstIndex(of: "c") {
    let distance = str.distance(from: str.startIndex, to: index)
    // distance is 2
}

在思考方面，这可能被称为反转。你会发现世界是圆的而不是平的。“你真的不需要知道角色的索引来处理它。”作为一名C程序员，我发现这也很难接受! 你的行“let index = letters.characters.indexOf("c")!”本身就足够了。 例如，要去掉c，你可以用…(操场粘贴)

    var letters = "abcdefg"
  //let index = letters.rangeOfString("c")!.startIndex //is the same as
    let index = letters.characters.indexOf("c")!
    range = letters.characters.indexOf("c")!...letters.characters.indexOf("c")!
    letters.removeRange(range)
    letters

然而，如果你想要一个索引，你需要返回一个实际的index而不是Int值，因为Int值对于任何实际使用都需要额外的步骤。这些扩展返回一个索引，一个特定字符的计数，以及这个游乐场插件代码将演示的范围。

extension String
{
    public func firstIndexOfCharacter(aCharacter: Character) -> String.CharacterView.Index? {

        for index in self.characters.indices {
            if self[index] == aCharacter {
                return index
            }

        }
        return nil
    }

    public func returnCountOfThisCharacterInString(aCharacter: Character) -> Int? {

        var count = 0
        for letters in self.characters{

            if aCharacter == letters{

                count++
            }
        }
        return count
    }


    public func rangeToCharacterFromStart(aCharacter: Character) -> Range<Index>? {

        for index in self.characters.indices {
            if self[index] == aCharacter {
                let range = self.startIndex...index
                return range
            }

        }
        return nil
    }

}



var MyLittleString = "MyVery:important String"

var theIndex = MyLittleString.firstIndexOfCharacter(":")

var countOfColons = MyLittleString.returnCountOfThisCharacterInString(":")

var theCharacterAtIndex:Character = MyLittleString[theIndex!]

var theRange = MyLittleString.rangeToCharacterFromStart(":")
MyLittleString.removeRange(theRange!)

如果你只需要一个字符的索引，最简单，快速的解决方案(正如Pascal已经指出的那样)是:

let index = string.characters.index(of: ".")
let intIndex = string.distance(from: string.startIndex, to: index)

    // Using Swift 4, the code below works.
    // The problem is that String.index is a struct. Use dot notation to grab the integer part of it that you want: ".encodedOffset"
    let strx = "0123456789ABCDEF"
    let si = strx.index(of: "A")
    let i = si?.encodedOffset       // i will be an Int. You need "?" because it might be nil, no such character found.

    if i != nil {                   // You MUST deal with the optional, unwrap it only if not nil.
        print("i = ",i)
        print("i = ",i!)            // "!" str1ps off "optional" specification (unwraps i).
            // or
        let ii = i!
        print("ii = ",ii)

    }
    // Good luck.