在Swift 4中：

Question

我曾经在JavaScript中这样做过：

var domains = "abcde".substring(0, "abcde".indexOf("cd")) // Returns "ab"

斯威夫特没有这个功能，如何做类似的事情呢？

Answer 1

编辑/更新：

Xcode 10.2.x•Swift 5或更高版本

extension StringProtocol { // for Swift 4.x syntax you will needed also to constrain the collection Index to String Index - `extension StringProtocol where Index == String.Index`
    func index(of string: Self, options: String.CompareOptions = []) -> Index? {
        return range(of: string, options: options)?.lowerBound
    }
    func endIndex(of string: Self, options: String.CompareOptions = []) -> Index? {
        return range(of: string, options: options)?.upperBound
    }
    func indexes(of string: Self, options: String.CompareOptions = []) -> [Index] {
        var result: [Index] = []
        var startIndex = self.startIndex
        while startIndex < endIndex,
            let range = self[startIndex...].range(of: string, options: options) {
                result.append(range.lowerBound)
                startIndex = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
    func ranges(of string: Self, options: String.CompareOptions = []) -> [Range<Index>] {
        var result: [Range<Index>] = []
        var startIndex = self.startIndex
        while startIndex < endIndex,
            let range = self[startIndex...].range(of: string, options: options) {
                result.append(range)
                startIndex = range.lowerBound < range.upperBound ? range.upperBound :
                    index(range.lowerBound, offsetBy: 1, limitedBy: endIndex) ?? endIndex
        }
        return result
    }
}

<强>用法：

let str = "abcde"
if let index = str.index(of: "cd") {
    let substring = str[..<index]   // ab
    let string = String(substring)
    print(string)  // "ab\n"
}

let str = "Hello, playground, playground, playground"
str.index(of: "play")      // 7
str.endIndex(of: "play")   // 11
str.indexes(of: "play")    // [7, 19, 31]
str.ranges(of: "play")     // [{lowerBound 7, upperBound 11}, {lowerBound 19, upperBound 23}, {lowerBound 31, upperBound 35}]

不区分大小写的样本

let query = "Play"
let ranges = str.ranges(of: query, options: .caseInsensitive)
let matches = ranges.map { str[$0] }   //
print(matches)  // ["play", "play", "play"]

正则表达式样本

let query = "play"
let escapedQuery = NSRegularExpression.escapedPattern(for: query)
let pattern = "\\b\(escapedQuery)\\w+"  // matches any word that starts with "play" prefix

let ranges = str.ranges(of: pattern, options: .regularExpression)
let matches = ranges.map { str[$0] }

print(matches) //  ["playground", "playground", "playground"]

Answer 2

测试Swift 4.2 / 4.1 / 4.0 / 3.0

使用 String[Range<String.Index>] 下标，您可以获得子字符串。您需要开始索引和最后一个索引来创建范围，您可以按照下面的方式执行

let str = "abcde"
if let range = str.range(of: "cd") {
  let substring = str[..<range.lowerBound] // or str[str.startIndex..<range.lowerBound]
  print(substring)  // Prints ab
}
else {
  print("String not present")
}

如果未定义此运算符..<的起始索引，则采用起始索引。您也可以使用str[str.startIndex..<range.lowerBound]代替str[..<range.lowerBound]

Answer 3

在Swift 4中：

获取字符串中的字符索引：

let str = "abcdefghabcd"
if let index = str.index(of: "b") {
   print(index) // Index(_compoundOffset: 4, _cache: Swift.String.Index._Cache.character(1))
}

使用Swift 4从创建SubString（前缀和后缀）：

let str : String = "ilike"
for i in 0...str.count {
    let index = str.index(str.startIndex, offsetBy: i) // String.Index
    let prefix = str[..<index] // String.SubSequence
    let suffix = str[index...] // String.SubSequence
    print("prefix \(prefix), suffix : \(suffix)")
}

输出

prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix :

如果要生成2个索引之间的子字符串，请使用：

let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex

Answer 4

可以在Swift中执行此操作，但需要更多行，这是一个函数indexOf()执行预期的操作：

func indexOf(source: String, substring: String) -> Int? {
    let maxIndex = source.characters.count - substring.characters.count
    for index in 0...maxIndex {
        let rangeSubstring = source.startIndex.advancedBy(index)..<source.startIndex.advancedBy(index + substring.characters.count)
        if source.substringWithRange(rangeSubstring) == substring {
            return index
        }
    }
    return nil
}

var str = "abcde"
if let indexOfCD = indexOf(str, substring: "cd") {
    let distance = str.startIndex.advancedBy(indexOfCD)
    print(str.substringToIndex(distance)) // Returns "ab"
}

此功能未进行优化，但它可以完成短字符串的工作。

Answer 5

在Swift版本3中，String没有像 -

这样的函数

str.index(of: String)

如果子字符串需要索引，则获取范围的方法之一。我们在返回范围的字符串中有以下函数 -

str.range(of: <String>)
str.rangeOfCharacter(from: <CharacterSet>)
str.range(of: <String>, options: <String.CompareOptions>, range: <Range<String.Index>?>, locale: <Locale?>)

例如，在str

中查找第一次出现的游戏的索引

var str = "play play play"
var range = str.range(of: "play")
range?.lowerBound //Result : 0
range?.upperBound //Result : 4

注意：范围是可选的。如果它无法找到String，它将使它为零。例如

var str = "play play play"
var range = str.range(of: "zoo") //Result : nil
range?.lowerBound //Result : nil
range?.upperBound //Result : nil

Answer 6

快速5

查找子字符串的索引

let str = "abcdecd"
if let range: Range<String.Index> = str.range(of: "cd") {
    let index: Int = str.distance(from: str.startIndex, to: range.lowerBound)
    print("index: ", index) //index: 2
}
else {
    print("substring not found")
}

查找字符索引

let str = "abcdecd"
if let firstIndex = str.firstIndex(of: "c") {
    let index = str.distance(from: str.startIndex, to: firstIndex)
    print("index: ", index)   //index: 2
}
else {
    print("symbol not found")
}

Answer 7

您是否考虑过使用NSRange？

if let range = mainString.range(of: mySubString) {
  //...
}

Answer 8

Leo Dabus的答案很好。这是我基于他使用compactMap来避免Index out of range错误的答案。

Swift 5.1

extension StringProtocol {
    func ranges(of targetString: Self, options: String.CompareOptions = [], locale: Locale? = nil) -> [Range<String.Index>] {

        let result: [Range<String.Index>] = self.indices.compactMap { startIndex in
            let targetStringEndIndex = index(startIndex, offsetBy: targetString.count, limitedBy: endIndex) ?? endIndex
            return range(of: targetString, options: options, range: startIndex..<targetStringEndIndex, locale: locale)
        }
        return result
    }
}

// Usage
let str = "Hello, playground, playground, playground"
let ranges = str.ranges(of: "play")
ranges.forEach {
    print("[\($0.lowerBound.utf16Offset(in: str)), \($0.upperBound.utf16Offset(in: str))]")
}

// result - [7, 11], [19, 23], [31, 35]

Answer 9

这里有三个密切相关的问题：

所有子字符串查找方法都在Cocoa NSString世界中完成（基础）
基金会NSRange与Swift Range不匹配;前者使用起点和长度，后者使用终点
一般情况下，使用String.Index索引Swift字符，而不使用Int，但使用Int对基础字符进行索引，并且它们之间没有简单的直接转换（因为Foundation）和斯威夫特对角色的构成有不同的看法。

考虑到这一切，让我们考虑如何写作：

func substring(of s: String, from:Int, toSubstring s2 : String) -> Substring? {
    // ?
}

必须使用String Foundation方法在s2中搜索子字符串s。结果范围返回给我们，不作为NSRange（即使这是一个Foundation方法），但是作为String.Index的范围（包含在Optional中，以防我们没有＆＃ 39;根本找不到子串。但是，另一个数字from是一个Int。因此，我们不能形成任何涉及它们的范围。

但我们不必！我们所要做的就是使用一个String.Index的方法切掉原始字符串的 end ，然后使用原始字符串的 start 切掉采用Int的方法。幸运的是，存在这样的方法！像这样：

func substring(of s: String, from:Int, toSubstring s2 : String) -> Substring? {
    guard let r = s.range(of:s2) else {return nil}
    var s = s.prefix(upTo:r.lowerBound)
    s = s.dropFirst(from)
    return s
}

或者，如果您希望能够将此方法直接应用于字符串，就像这样...

let output = "abcde".substring(from:0, toSubstring:"cd")

...然后将其作为String的扩展名：

extension String {
    func substring(from:Int, toSubstring s2 : String) -> Substring? {
        guard let r = self.range(of:s2) else {return nil}
        var s = self.prefix(upTo:r.lowerBound)
        s = s.dropFirst(from)
        return s
    }
}

Answer 10

快捷键5

    extension String {
    enum SearchDirection {
        case first, last
    }
    func characterIndex(of character: Character, direction: String.SearchDirection) -> Int? {
        let fn = direction == .first ? firstIndex : lastIndex
        if let stringIndex: String.Index = fn(character) {
            let index: Int = distance(from: startIndex, to: stringIndex)
            return index
        }  else {
            return nil
        }
    }
}

测试：

 func testFirstIndex() {
        let res = ".".characterIndex(of: ".", direction: .first)
        XCTAssert(res == 0)
    }
    func testFirstIndex1() {
        let res = "12345678900.".characterIndex(of: "0", direction: .first)
        XCTAssert(res == 9)
    }
    func testFirstIndex2() {
        let res = ".".characterIndex(of: ".", direction: .last)
        XCTAssert(res == 0)
    }
    func testFirstIndex3() {
        let res = "12345678900.".characterIndex(of: "0", direction: .last)
        XCTAssert(res == 10)
    }

Answer 11

快捷键5

   let alphabat = "abcdefghijklmnopqrstuvwxyz"

    var index: Int = 0
    
    if let range: Range<String.Index> = alphabat.range(of: "c") {
         index = alphabat.distance(from: alphabat.startIndex, to: range.lowerBound)
        print("index: ", index) //index: 2
    }

使用Swift的字符串中子字符串的索引

11 个答案:

在Swift 4中：

获取字符串中的字符索引：

使用Swift 4从创建SubString（前缀和后缀）：

输出

如果要生成2个索引之间的子字符串，请使用：

Swift 5.1