我找到了swift2的解决方案:

var str = "abcdefghi"
let indexForCharacterInString = str.characters.indexOf("c") //returns 2

    // 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.

let mystring:String = "indeep";
let findCharacter:Character = "d";

if (mystring.characters.contains(findCharacter))
{
    let position = mystring.characters.indexOf(findCharacter);
    NSLog("Position of c is \(mystring.startIndex.distanceTo(position!))")

}
else
{
    NSLog("Position of c is not found");
}

Swift 4完整解决方案:

OffsetIndexableCollection(使用Int索引的字符串)

https://github.com/frogcjn/OffsetIndexableCollection-String-Int-Indexable-

let a = "01234"

print(a[0]) // 0
print(a[0...4]) // 01234
print(a[...]) // 01234

print(a[..<2]) // 01
print(a[...2]) // 012
print(a[2...]) // 234
print(a[2...3]) // 23
print(a[2...2]) // 2

if let number = a.index(of: "1") {
    print(number) // 1
    print(a[number...]) // 1234
}

if let number = a.index(where: { $0 > "1" }) {
    print(number) // 2
}

Swift 3.0让这个更加冗长:

let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.index(of: needle) {
    let pos = string.characters.distance(from: string.startIndex, to: idx)
    print("Found \(needle) at position \(pos)")
}
else {
    print("Not found")
}

扩展:

extension String {
    public func index(of char: Character) -> Int? {
        if let idx = characters.index(of: char) {
            return characters.distance(from: startIndex, to: idx)
        }
        return nil
    }
}

在Swift 2.0中，这变得更加容易:

let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.indexOf(needle) {
    let pos = string.startIndex.distanceTo(idx)
    print("Found \(needle) at position \(pos)")
}
else {
    print("Not found")
}

扩展:

extension String {
    public func indexOfCharacter(char: Character) -> Int? {
        if let idx = self.characters.indexOf(char) {
            return self.startIndex.distanceTo(idx)
        }
        return nil
    }
}

斯威夫特1。x实现:

对于纯Swift解决方案，可以使用:

let string = "Hello.World"
let needle: Character = "."
if let idx = find(string, needle) {
    let pos = distance(string.startIndex, idx)
    println("Found \(needle) at position \(pos)")
}
else {
    println("Not found")
}

作为String的扩展:

extension String {
    public func indexOfCharacter(char: Character) -> Int? {
        if let idx = find(self, char) {
            return distance(self.startIndex, idx)
        }
        return nil
    }
}

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