我怎样才能得到字符串的第n个字符?我尝试了括号([])访问器,没有运气。

var string = "Hello, world!"

var firstChar = string[0] // Throws error

错误:'下标'是不可用的:不能下标String与Int,请参阅文档注释讨论


当前回答

如果你看到不能下标一个类型为'String'的值…使用这个扩展:

斯威夫特3

extension String {
    subscript (i: Int) -> Character {
        return self[self.characters.index(self.startIndex, offsetBy: i)]
    }

    subscript (i: Int) -> String {
        return String(self[i] as Character)
    }

    subscript (r: Range<Int>) -> String {
        let start = index(startIndex, offsetBy: r.lowerBound)
        let end = index(startIndex, offsetBy: r.upperBound)
        return self[start..<end]
    }

    subscript (r: ClosedRange<Int>) -> String {
        let start = index(startIndex, offsetBy: r.lowerBound)
        let end = index(startIndex, offsetBy: r.upperBound)
        return self[start...end]
    }
}

斯威夫特2.3

extension String {
    subscript(integerIndex: Int) -> Character {
        let index = advance(startIndex, integerIndex)
        return self[index]
    }

    subscript(integerRange: Range<Int>) -> String {
        let start = advance(startIndex, integerRange.startIndex)
        let end = advance(startIndex, integerRange.endIndex)
        let range = start..<end
        return self[range]
    }
}

来源:http://oleb.net/blog/2014/07/swift-strings/

其他回答

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 ???

斯威夫特5.2

let str = "abcdef"
str[1 ..< 3] // returns "bc"
str[5] // returns "f"
str[80] // returns ""
str.substring(fromIndex: 3) // returns "def"
str.substring(toIndex: str.length - 2) // returns "abcd"

你需要将这个String扩展添加到你的项目中(它已经完全测试过了):

extension String {

    var length: Int {
        return 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])
    }
}

尽管Swift总是有开箱即用的解决方案来解决这个问题(没有字符串扩展,我在下面提供),我仍然强烈建议使用扩展。为什么?因为它为我从早期版本的Swift中节省了数十个小时的痛苦迁移,在早期版本中,String的语法几乎每次发布都要更改,但我所需要做的只是更新扩展的实现,而不是重构整个项目。做出你的选择。

let str = "Hello, world!"
let index = str.index(str.startIndex, offsetBy: 4)
str[index] // returns Character 'o'

let endIndex = str.index(str.endIndex, offsetBy:-2)
str[index ..< endIndex] // returns String "o, worl"

String(str.suffix(from: index)) // returns String "o, world!"
String(str.prefix(upTo: index)) // returns String "Hell"

斯威夫特4.2

这个答案是理想的,因为它在一个扩展中扩展了String及其所有子序列(Substring)

public extension StringProtocol {
    
    public subscript (i: Int) -> Element {
        return self[index(startIndex, offsetBy: i)]
    }

    public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start...end]
    }
    
    public subscript (bounds: CountableRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start..<end]
    }
    
    public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex..<end]
    }
    
    public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex...end]
    }
    
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        return self[start..<endIndex]
    }
}

使用

var str = "Hello, playground"

print(str[5...][...5][0])
// Prints ","

Swift 2.0的Xcode 7通用种子

var text = "Hello, world!"

let firstChar = text[text.startIndex.advancedBy(0)] // "H"

对于第n个字符,将0替换为n-1。

编辑:Swift 3.0

text[text.index(text.startIndex, offsetBy: 0)]

注意,有更简单的方法来获取字符串中的某些字符

例如:let firstChar = text.characters.first

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)