我习惯使用JavaScript来做这件事:
var domains = "abcde".substring(0, "abcde".indexOf("cd")) // Returns "ab"
Swift 没有这个函数,如何做类似的事情?
编辑/更新:
Xcode 11.4 • Swift 5.2或更高版本
import Foundation
extension StringProtocol {
func index<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
range(of: string, options: options)?.lowerBound
}
func endIndex<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> Index? {
range(of: string, options: options)?.upperBound
}
func indices<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Index] {
ranges(of: string, options: options).map(\.lowerBound)
}
func ranges<S: StringProtocol>(of string: S, 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
}
}
使用方法:
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.indices(of: "play") // [7, 19, 31]
str.ranges(of: "play") // [{lowerBound 7, upperBound 11}, {lowerBound 19, upperBound 23}, {lowerBound 31, upperBound 35}]
不区分大小写的示例
let query = "Play"
let ranges = str.ranges(of: query, options: .caseInsensitive)
let matches = ranges.map { str[$0] } //
print(matches) // ["play", "play", "play"]
正则表达式示例
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"]
"ab".indexOf("a")
和 "ab".indexOf("c")
都会返回 0
。 - Grahamfunc indexOf(string: String) -> String.Index? {
return range(of: string, options: .literal, range: nil, locale: nil)?.lowerBound
} }
- gammachill使用 String[Range<String.Index>]
下标可以获取子字符串。你需要提供起始索引和结束索引来创建范围,如下所示:
使用 String[Range<String.Index>]
下标可以获取子字符串。您需要提供起始索引和结束索引以创建范围,方法如下:
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")
}
如果您不定义起始索引,这个运算符 ..<
会采用起始索引。您也可以使用 str[str.startIndex..<range.lowerBound]
替代 str[..<range.lowerBound]
Swift 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")
}
查找字符的索引
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")
}
let str = "abcdefghabcd"
if let index = str.index(of: "b") {
print(index) // Index(_compoundOffset: 4, _cache: Swift.String.Index._Cache.character(1))
}
let str = "Hello, World"
let prefix = str.prefix(5) // "Hello"
prefix(_ maxLength: Int)
方法来获取字符串的前五个字符。
后缀:let str = "Hello, World"
let suffix = str.suffix(5) // "World"
suffix(_ maxLength: Int)
方法来获取字符串的最后五个字符。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)")
}
prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix :
let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex
string[startIndex...endIndex]
会导致崩溃。顺便说一下,在 Swift 5 或更高版本中,你可以使用 PartialRangeFrom
下标来实现 let substring1 = str[str.startIndex...]
。 - Leo DabusindexOf()
,可以完成所需功能: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"
}
这个函数并没有进行优化,但是对于短字符串来说它可以完成任务。
utils.swift
的类中添加一个extension String
,这将需要对所有其他类都可用。 - WestCoastProjectsO(N^2)
的时间复杂度..? - WestCoastProjectsSwift 5
let alphabet = "abcdefghijklmnopqrstuvwxyz"
var index: Int = 0
if let range: Range<String.Index> = alphabet.range(of: "c") {
index = alphabet.distance(from: alphabet.startIndex, to: range.lowerBound)
print("index: ", index) //index: 2
}
在Cocoa NSString世界(Foundation)中,所有子字符串查找方法都已经过时了
Foundation NSRange与Swift Range不匹配;前者使用起始位置和长度,后者使用端点
一般来说,Swift字符使用String.Index进行索引,而不是Int,但Foundation字符使用Int进行索引,并且它们之间没有简单的直接转换(因为Foundation和Swift对什么构成字符有不同的想法)
func substring(of s: String, from:Int, toSubstring s2 : String) -> Substring? {
// ?
}
但是我们不必!我们所要做的就是使用一个以String.Index为参数的方法切掉原始字符串的结尾,并使用一个以Int为参数的方法切掉原始字符串的开头。幸运的是,这样的方法是存在的!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
}
let output = "abcde".substring(from:0, toSubstring:"cd")
...然后将其作为String的扩展:
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
}
}
dropFirst
,但还没有查看它的实现方式。我们如何将最终返回的 Substring
提取为一个 String
?我看到一些超长的帖子只是关于这个问题的... - WestCoastProjectsas! String
时,出现警告“从 Substring 转换为 String 的强制转换始终失败”。 - WestCoastProjectsjvm
上“cast”/“coercion”术语的区别。 - WestCoastProjectsSwift 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)
}
Leo Dabus的回答很好。这里是基于他的回答,使用compactMap
来避免Index out of range
错误的方法。
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]
str.index(of: String)
如果需要获取子字符串的索引,其中一种方法是获取范围。字符串中有以下函数返回范围 -
str.range(of: <String>)
str.rangeOfCharacter(from: <CharacterSet>)
str.range(of: <String>, options: <String.CompareOptions>, range: <Range<String.Index>?>, locale: <Locale?>)
var str = "play play play"
var range = str.range(of: "play")
range?.lowerBound //Result : 0
range?.upperBound //Result : 4
注意:range是可选的。如果无法找到字符串,它将使其为nil。例如:
var str = "play play play"
var range = str.range(of: "zoo") //Result : nil
range?.lowerBound //Result : nil
range?.upperBound //Result : nil