允许负指数

它总是有用的，不必总是写string[string]。长度- 1]用于在使用下标扩展名时获取最后一个字符。这(Swift 3)扩展允许负索引，范围和CountableClosedRange。

extension String {
    var count: Int { return self.characters.count }

    subscript (i: Int) -> Character {
        // wraps out of bounds indices
        let j = i % self.count
        // wraps negative indices
        let x = j < 0 ? j + self.count : j

        // quick exit for first
        guard x != 0 else {
            return self.characters.first!
        }

        // quick exit for last
        guard x != count - 1 else {
            return self.characters.last!
        }

        return self[self.index(self.startIndex, offsetBy: x)]
    }

    subscript (r: Range<Int>) -> String {
        let lb = r.lowerBound
        let ub = r.upperBound

        // quick exit for one character
        guard lb != ub else { return String(self[lb]) }

        return self[self.index(self.startIndex, offsetBy: lb)..<self.index(self.startIndex, offsetBy: ub)]
    }

    subscript (r: CountableClosedRange<Int>) -> String {
        return self[r.lowerBound..<r.upperBound + 1]
    }
}

如何使用:

var text = "Hello World"

text[-1]    // d
text[2]     // l
text[12]    // e
text[0...4] // Hello
text[0..<4] // Hell

对于更彻底的程序员:在这个扩展中包括一个防止空字符串的保护

subscript (i: Int) -> Character {
    guard self.count != 0 else { return '' }
    ...
}

subscript (r: Range<Int>) -> String {
    guard self.count != 0 else { return "" }
    ...
}

斯威夫特4

let str = "My String"

索引处的字符串

let index = str.index(str.startIndex, offsetBy: 3)
String(str[index])    // "S"

子字符串

let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
String(str[startIndex...endIndex])     // "Strin"

前n个字符

let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[..<startIndex])    // "My "

最后n个字符

let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[startIndex...])    // "String"

Swift 2和3

str = "My String"

**字符串索引**

斯威夫特2

let charAtIndex = String(str[str.startIndex.advancedBy(3)])  // charAtIndex = "S"

斯威夫特3

str[str.index(str.startIndex, offsetBy: 3)]

子字符串fromIndex toIndex

斯威夫特2

let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)] // subStr = "Strin"

斯威夫特3

str[str.index(str.startIndex, offsetBy: 3)...str.index(str.startIndex, offsetBy: 7)]

前n个字符

let first2Chars = String(str.characters.prefix(2)) // first2Chars = "My"

最后n个字符

let last3Chars = String(str.characters.suffix(3)) // last3Chars = "ing"

注意:请参阅Leo Dabus关于正确实现Swift 4和Swift 5的回答。

Swift 4或更高版本

Substring类型是在Swift 4中引入的，用于生成子字符串 通过与原始字符串共享存储，更快更有效，这就是下标函数应该返回的。

在这里试试吧

extension StringProtocol {
    subscript(offset: Int) -> Character { self[index(startIndex, offsetBy: offset)] }
    subscript(range: Range<Int>) -> SubSequence {
        let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
        return self[startIndex..<index(startIndex, offsetBy: range.count)]
    }
    subscript(range: ClosedRange<Int>) -> SubSequence {
        let startIndex = index(self.startIndex, offsetBy: range.lowerBound)
        return self[startIndex..<index(startIndex, offsetBy: range.count)]
    }
    subscript(range: PartialRangeFrom<Int>) -> SubSequence { self[index(startIndex, offsetBy: range.lowerBound)...] }
    subscript(range: PartialRangeThrough<Int>) -> SubSequence { self[...index(startIndex, offsetBy: range.upperBound)] }
    subscript(range: PartialRangeUpTo<Int>) -> SubSequence { self[..<index(startIndex, offsetBy: range.upperBound)] }
}

要将子字符串转换为字符串，您可以简单地 做字符串(字符串[0..2])，但你应该只做如果 您计划保留子字符串。否则，就更多了 有效地保持它为Substring。

It would be great if someone could figure out a good way to merge these two extensions into one. I tried extending StringProtocol without success, because the index method does not exist there. Note: This answer has been already edited, it is properly implemented and now works for substrings as well. Just make sure to use a valid range to avoid crashing when subscripting your StringProtocol type. For subscripting with a range that won't crash with out of range values you can use this implementation

为什么这不是内置的?

错误消息显示“请参阅文档注释以进行讨论”。Apple在文件unavailablestringapi .swift中提供了以下解释:

Subscripting strings with integers is not available. The concept of "the ith character in a string" has different interpretations in different libraries and system components. The correct interpretation should be selected according to the use case and the APIs involved, so String cannot be subscripted with an integer. Swift provides several different ways to access the character data stored inside strings. String.utf8 is a collection of UTF-8 code units in the string. Use this API when converting the string to UTF-8. Most POSIX APIs process strings in terms of UTF-8 code units. String.utf16 is a collection of UTF-16 code units in string. Most Cocoa and Cocoa touch APIs process strings in terms of UTF-16 code units. For example, instances of NSRange used with NSAttributedString and NSRegularExpression store substring offsets and lengths in terms of UTF-16 code units. String.unicodeScalars is a collection of Unicode scalars. Use this API when you are performing low-level manipulation of character data. String.characters is a collection of extended grapheme clusters, which are an approximation of user-perceived characters. Note that when processing strings that contain human-readable text, character-by-character processing should be avoided to the largest extent possible. Use high-level locale-sensitive Unicode algorithms instead, for example, String.localizedStandardCompare(), String.localizedLowercaseString, String.localizedStandardRangeOfString() etc.

Swift 3:另一个解决方案(在操场测试)

extension String {
    func substr(_ start:Int, length:Int=0) -> String? {
        guard start > -1 else {
            return nil
        }

        let count = self.characters.count - 1

        guard start <= count else {
            return nil
        }

        let startOffset = max(0, start)
        let endOffset = length > 0 ? min(count, startOffset + length - 1) : count

        return self[self.index(self.startIndex, offsetBy: startOffset)...self.index(self.startIndex, offsetBy: endOffset)]
    }
}

用法:

let txt = "12345"

txt.substr(-1) //nil
txt.substr(0) //"12345"
txt.substr(0, length: 0) //"12345"
txt.substr(1) //"2345"
txt.substr(2) //"345"
txt.substr(3) //"45"
txt.substr(4) //"5"
txt.substr(6) //nil
txt.substr(0, length: 1) //"1"
txt.substr(1, length: 1) //"2"
txt.substr(2, length: 1) //"3"
txt.substr(3, length: 1) //"4"
txt.substr(3, length: 2) //"45"
txt.substr(3, length: 3) //"45"
txt.substr(4, length: 1) //"5"
txt.substr(4, length: 2) //"5"
txt.substr(5, length: 1) //nil
txt.substr(5, length: -1) //nil
txt.substr(-1, length: -1) //nil

斯威夫特5.3

我觉得这很优雅。“Hacking with Swift”的Paul Hudson提出了以下解决方案:

@available (macOS 10.15, * )
extension String {
    subscript(idx: Int) -> String {
        String(self[index(startIndex, offsetBy: idx)])
    }
}

然后，要从字符串中获取一个字符，你只需做:

var string = "Hello, world!"

var firstChar = string[0] // No error, returns "H" as a String

注意:我只是想补充，这将返回一个字符串指出在评论。我认为这对Swift用户来说可能是意想不到的，但我经常需要一个字符串来直接在我的代码中使用，而不是字符类型，所以它确实简化了我的代码，避免了后来从字符到字符串的转换。

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 ","