Swift：如何获取字符从开始到最后一个索引的子string

我想学习最好的/最简单的方法，把一个string转换成另一个string，但只有一个子集，从一开始就进入一个字符的最后一个索引。

例如，将“www.stackoverflow.com”转换为“www.stackoverflow”。 什么代码片段会做到这一点，而且是最快的？ （我希望这不会引起辩论，但我无法find如何处理Swift中的子string的好教训。

Answer 1

只是后退

最好的方法是将substringToIndex合并到endIndex属性和advance全局函数中。

 var string1 = "www.stackoverflow.com" var index1 = advance(string1.endIndex, -4) var substring1 = string1.substringToIndex(index1) 

寻找从后面开始的string

使用rangeOfString并将options设置为.BackwardsSearch

 var string2 = "www.stackoverflow.com" var index2 = string2.rangeOfString(".", options: .BackwardsSearch)?.startIndex var substring2 = string2.substringToIndex(index2!) 

没有扩展，纯粹的惯用Swift

Swift 2.0

advance现在是Index的一部分，被称为advancedBy 。 你这样做：

 var string1 = "www.stackoverflow.com" var index1 = string1.endIndex.advancedBy(-4) var substring1 = string1.substringToIndex(index1) 

Swift 3.0

你不能在String上调用advancedBy ，因为它具有可变大小的元素。 你必须使用index(_, offsetBy:) 。

 var string1 = "www.stackoverflow.com" var index1 = string1.index(string1.endIndex, offsetBy: -4) var substring1 = string1.substring(to: index1) 

很多东西都被重新命名了。 这些情况是用camelCase写的， startIndex变成了lowerBound 。

 var string2 = "www.stackoverflow.com" var index2 = string2.range(of: ".", options: .backwards)?.lowerBound var substring2 = string2.substring(to: index2!) 

另外，我不会推荐强制展开index2 。 您可以使用可选的绑定或map 。 就个人而言，我更喜欢使用map ：

 var substring3 = index2.map(string2.substring(to:)) 

斯威夫特4

Swift 3版本仍然有效，但现在可以使用索引范围的下标了：

 let string1 = "www.stackoverflow.com" let index1 = string1.index(string1.endIndex, offsetBy: -4) let substring1 = string1[string1.startIndex..<index1] 

第二种方法保持不变：

 let string2 = "www.stackoverflow.com" let index2 = string2.range(of: ".", options: .backwards)?.lowerBound let substring3 = index2.map(string2.substring(to:)) 

Answer 2

Swift 3，XCode 8

由于advancedBy(Int)不在使用String的方法index(String.Index, Int) 。 看看这个String扩展名：

 public extension String { func substring(_ r: Range<Int>) -> String { let fromIndex = self.index(self.startIndex, offsetBy: r.lowerBound) let toIndex = self.index(self.startIndex, offsetBy: r.upperBound) return self.substring(with: Range<String.Index>(uncheckedBounds: (lower: fromIndex, upper: toIndex))) } } 

例：

 let str = "0😀123456789" print(str.substring(0..<3)) // prints 0😀1 

以下是可能有用的更多方法：

 public extension String { //right is the first encountered string after left func between(_ left: String, _ right: String) -> String? { guard let leftRange = range(of: left), let rightRange = range(of: right, options: .backwards) , left != right && leftRange.upperBound < rightRange.lowerBound else { return nil } let sub = self.substring(from: leftRange.upperBound) let closestToLeftRange = sub.range(of: right)! return sub.substring(to: closestToLeftRange.lowerBound) } var length: Int { get { return self.characters.count } } func substring(to : Int) -> String? { if (to >= length) { return nil } let toIndex = self.index(self.startIndex, offsetBy: to) return self.substring(to: toIndex) } func substring(from : Int) -> String? { if (from >= length) { return nil } let fromIndex = self.index(self.startIndex, offsetBy: from) return self.substring(from: fromIndex) } func substring(_ r: Range<Int>) -> String { let fromIndex = self.index(self.startIndex, offsetBy: r.lowerBound) let toIndex = self.index(self.startIndex, offsetBy: r.upperBound) return self.substring(with: Range<String.Index>(uncheckedBounds: (lower: fromIndex, upper: toIndex))) } func character(_ at: Int) -> Character { return self[self.index(self.startIndex, offsetBy: at)] } } 

PS开发人员不得不处理String.Index而不是纯Int ，这真是太奇怪了。 为什么我们应该打扰内部的String机制，而不是简单的substring()方法？

Answer 3

我会用下标（ s[start..<end] ）来做：

Swift 2

 let s = "www.stackoverflow.com" let start = s.startIndex let end = s.endIndex.advancedBy(-4) let substring = s[start..<end] // www.stackoverflow 

Swift 3

 let s = "www.stackoverflow.com" let start = s.startIndex let end = s.index(s.endIndex, offsetBy: -4) let substring = s[start..<end] // www.stackoverflow 

Answer 4

例如，将“www.stackoverflow.com”转换为“www.stackoverflow”。 什么代码片段会做到这一点，而且是最快的？

它能更像Swift吗？

 let myString = "www.stackoverflow.com".stringByDeletingPathExtension // "www.stackoverflow" 

更新： Xcode 7.2.1•Swift 2.1.1

 extension String { var nsValue: NSString { return self } } let myString = "www.stackoverflow.com".nsValue.stringByDeletingPathExtension // "www.stackoverflow" 

Answer 5

这是我怎么做的。 你可以用同样的方法做，也可以使用这个代码来获得想法。

 let s = "www.stackoverflow.com" s.substringWithRange(0..<s.lastIndexOf(".")) 

这里是我使用的扩展：

 import Foundation extension String { var length: Int { get { return countElements(self) } } func indexOf(target: String) -> Int { var range = self.rangeOfString(target) if let range = range { return distance(self.startIndex, range.startIndex) } else { return -1 } } func indexOf(target: String, startIndex: Int) -> Int { var startRange = advance(self.startIndex, startIndex) var range = self.rangeOfString(target, options: NSStringCompareOptions.LiteralSearch, range: Range<String.Index>(start: startRange, end: self.endIndex)) if let range = range { return distance(self.startIndex, range.startIndex) } else { return -1 } } func lastIndexOf(target: String) -> Int { var index = -1 var stepIndex = self.indexOf(target) while stepIndex > -1 { index = stepIndex if stepIndex + target.length < self.length { stepIndex = indexOf(target, startIndex: stepIndex + target.length) } else { stepIndex = -1 } } return index } func substringWithRange(range:Range<Int>) -> String { let start = advance(self.startIndex, range.startIndex) let end = advance(self.startIndex, range.endIndex) return self.substringWithRange(start..<end) } } 

信用albertbori /通用Swiftstring扩展

一般来说，我是一个强大的扩展支持者，特别是像string操作，search和切片的需求。

Answer 6

您可以使用这些扩展名：

Swift 2.3

  extension String { func substringFromIndex(index: Int) -> String { if (index < 0 || index > self.characters.count) { print("index \(index) out of bounds") return "" } return self.substringFromIndex(self.startIndex.advancedBy(index)) } func substringToIndex(index: Int) -> String { if (index < 0 || index > self.characters.count) { print("index \(index) out of bounds") return "" } return self.substringToIndex(self.startIndex.advancedBy(index)) } func substringWithRange(start: Int, end: Int) -> String { if (start < 0 || start > self.characters.count) { print("start index \(start) out of bounds") return "" } else if end < 0 || end > self.characters.count { print("end index \(end) out of bounds") return "" } let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(end)) return self.substringWithRange(range) } func substringWithRange(start: Int, location: Int) -> String { if (start < 0 || start > self.characters.count) { print("start index \(start) out of bounds") return "" } else if location < 0 || start + location > self.characters.count { print("end index \(start + location) out of bounds") return "" } let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(start + location)) return self.substringWithRange(range) } } 

Swift 3

 extension String { func substring(from index: Int) -> String { if (index < 0 || index > self.characters.count) { print("index \(index) out of bounds") return "" } return self.substring(from: self.characters.index(self.startIndex, offsetBy: index)) } func substring(to index: Int) -> String { if (index < 0 || index > self.characters.count) { print("index \(index) out of bounds") return "" } return self.substring(to: self.characters.index(self.startIndex, offsetBy: index)) } func substring(start: Int, end: Int) -> String { if (start < 0 || start > self.characters.count) { print("start index \(start) out of bounds") return "" } else if end < 0 || end > self.characters.count { print("end index \(end) out of bounds") return "" } let startIndex = self.characters.index(self.startIndex, offsetBy: start) let endIndex = self.characters.index(self.startIndex, offsetBy: end) let range = startIndex..<endIndex return self.substring(with: range) } func substring(start: Int, location: Int) -> String { if (start < 0 || start > self.characters.count) { print("start index \(start) out of bounds") return "" } else if location < 0 || start + location > self.characters.count { print("end index \(start + location) out of bounds") return "" } let startIndex = self.characters.index(self.startIndex, offsetBy: start) let endIndex = self.characters.index(self.startIndex, offsetBy: start + location) let range = startIndex..<endIndex return self.substring(with: range) } } 

用法：

 let string = "www.stackoverflow.com" let substring = string.substringToIndex(string.characters.count-4) 

Answer 7

String具有内build的子stringfunction：

 extension String : Sliceable { subscript (subRange: Range<String.Index>) -> String { get } } 

如果你想要的是“去一个字符的第一个索引”，你可以使用内build的find()函数得到子string：

 var str = "www.stackexchange.com" str[str.startIndex ..< find(str, ".")!] // -> "www" 

要find最后的索引，我们可以实现findLast() 。

 /// Returns the last index where `value` appears in `domain` or `nil` if /// `value` is not found. /// /// Complexity: O(\ `countElements(domain)`\ ) func findLast<C: CollectionType where C.Generator.Element: Equatable>(domain: C, value: C.Generator.Element) -> C.Index? { var last:C.Index? = nil for i in domain.startIndex..<domain.endIndex { if domain[i] == value { last = i } } return last } let str = "www.stackexchange.com" let substring = map(findLast(str, ".")) { str[str.startIndex ..< $0] } // as String? // if "." is found, substring has some, otherwise `nil` 

---

添加：

也许， findLast BidirectionalIndexType专用版本更快：

 func findLast<C: CollectionType where C.Generator.Element: Equatable, C.Index: BidirectionalIndexType>(domain: C, value: C.Generator.Element) -> C.Index? { for i in lazy(domain.startIndex ..< domain.endIndex).reverse() { if domain[i] == value { return i } } return nil } 

Answer 8

你想从一个string的开始索引到其中一个字符的最后一个索引获得一个string的子string吗？ 如果是这样，你可以select下面的Swift 2.0+方法之一。

需要Foundation方法

获取包含一个字符最后一个索引的子string：

 import Foundation let string = "www.stackoverflow.com" if let rangeOfIndex = string.rangeOfCharacterFromSet(NSCharacterSet(charactersInString: "."), options: .BackwardsSearch) { print(string.substringToIndex(rangeOfIndex.endIndex)) } // prints "www.stackoverflow." 

获取不包含字符最后一个索引的子string：

 import Foundation let string = "www.stackoverflow.com" if let rangeOfIndex = string.rangeOfCharacterFromSet(NSCharacterSet(charactersInString: "."), options: .BackwardsSearch) { print(string.substringToIndex(rangeOfIndex.startIndex)) } // prints "www.stackoverflow" 

---

如果你需要重复这些操作，扩展String可以是一个很好的解决scheme：

 import Foundation extension String { func substringWithLastInstanceOf(character: Character) -> String? { if let rangeOfIndex = rangeOfCharacterFromSet(NSCharacterSet(charactersInString: String(character)), options: .BackwardsSearch) { return self.substringToIndex(rangeOfIndex.endIndex) } return nil } func substringWithoutLastInstanceOf(character: Character) -> String? { if let rangeOfIndex = rangeOfCharacterFromSet(NSCharacterSet(charactersInString: String(character)), options: .BackwardsSearch) { return self.substringToIndex(rangeOfIndex.startIndex) } return nil } } print("www.stackoverflow.com".substringWithLastInstanceOf(".")) print("www.stackoverflow.com".substringWithoutLastInstanceOf(".")) /* prints: Optional("www.stackoverflow.") Optional("www.stackoverflow") */ 

---

不需要Foundation

获取包含一个字符最后一个索引的子string：

 let string = "www.stackoverflow.com" if let reverseIndex = string.characters.reverse().indexOf(".") { print(string[string.startIndex ..< reverseIndex.base]) } // prints "www.stackoverflow." 

获取不包含字符最后一个索引的子string：

 let string = "www.stackoverflow.com" if let reverseIndex = string.characters.reverse().indexOf(".") { print(string[string.startIndex ..< reverseIndex.base.advancedBy(-1)]) } // prints "www.stackoverflow" 

---

如果你需要重复这些操作，扩展String可以是一个很好的解决scheme：

 extension String { func substringWithLastInstanceOf(character: Character) -> String? { if let reverseIndex = characters.reverse().indexOf(".") { return self[self.startIndex ..< reverseIndex.base] } return nil } func substringWithoutLastInstanceOf(character: Character) -> String? { if let reverseIndex = characters.reverse().indexOf(".") { return self[self.startIndex ..< reverseIndex.base.advancedBy(-1)] } return nil } } print("www.stackoverflow.com".substringWithLastInstanceOf(".")) print("www.stackoverflow.com".substringWithoutLastInstanceOf(".")) /* prints: Optional("www.stackoverflow.") Optional("www.stackoverflow") */ 

Answer 9

如果你知道索引，这里有一个简单的方法来得到一个子串：

 let s = "www.stackoverflow.com" let result = String(s.characters.prefix(17)) // "www.stackoverflow" 

如果您的索引超出string的长度，它不会崩溃应用程序：

 let s = "short" let result = String(s.characters.prefix(17)) // "short" 

这两个例子都是Swift 3准备好的 。

Answer 10

Swift 3：

 extension String { /// the length of the string var length: Int { return self.characters.count } /// Get substring, eg "ABCDE".substring(index: 2, length: 3) -> "CDE" /// /// - parameter index: the start index /// - parameter length: the length of the substring /// /// - returns: the substring public func substring(index: Int, length: Int) -> String { if self.length <= index { return "" } let leftIndex = self.index(self.startIndex, offsetBy: index) if self.length <= index + length { return self.substring(from: leftIndex) } let rightIndex = self.index(self.endIndex, offsetBy: -(self.length - index - length)) return self.substring(with: leftIndex..<rightIndex) } /// Get substring, eg -> "ABCDE".substring(left: 0, right: 2) -> "ABC" /// /// - parameter left: the start index /// - parameter right: the end index /// /// - returns: the substring public func substring(left: Int, right: Int) -> String { if length <= left { return "" } let leftIndex = self.index(self.startIndex, offsetBy: left) if length <= right { return self.substring(from: leftIndex) } else { let rightIndex = self.index(self.endIndex, offsetBy: -self.length + right + 1) return self.substring(with: leftIndex..<rightIndex) } } } 

你可以testing它如下：

  print("test: " + String("ABCDE".substring(index: 2, length: 3) == "CDE")) print("test: " + String("ABCDE".substring(index: 0, length: 3) == "ABC")) print("test: " + String("ABCDE".substring(index: 2, length: 1000) == "CDE")) print("test: " + String("ABCDE".substring(left: 0, right: 2) == "ABC")) print("test: " + String("ABCDE".substring(left: 1, right: 3) == "BCD")) print("test: " + String("ABCDE".substring(left: 3, right: 1000) == "DE")) 

Answer 11

 func substr(myString: String, start: Int, clen: Int)->String { var index2 = string1.startIndex.advancedBy(start) var substring2 = string1.substringFromIndex(index2) var index1 = substring2.startIndex.advancedBy(clen) var substring1 = substring2.substringToIndex(index1) return substring1 } substr(string1, start: 3, clen: 5) 

Answer 12

Swift 3

 let string = "www.stackoverflow.com" let first3Characters = String(string.characters.prefix(3)) // www let lastCharacters = string.characters.dropFirst(4) // stackoverflow.com (it would be a collection) //or by index let indexOfFouthCharacter = olNumber.index(olNumber.startIndex, offsetBy: 4) let first3Characters = olNumber.substring(to: indexOfFouthCharacter) // www let lastCharacters = olNumber.substring(from: indexOfFouthCharacter) // .stackoverflow.com 

好文章理解，为什么我们需要这个

Answer 13

我已经扩展了两个子string方法的string。 你可以使用from / to范围或from / length调用substring，如下所示：

 var bcd = "abcdef".substring(1,to:3) var cde = "abcdef".substring(2,to:-2) var cde = "abcdef".substring(2,length:3) extension String { public func substring(from:Int = 0, var to:Int = -1) -> String { if to < 0 { to = self.length + to } return self.substringWithRange(Range<String.Index>( start:self.startIndex.advancedBy(from), end:self.startIndex.advancedBy(to+1))) } public func substring(from:Int = 0, length:Int) -> String { return self.substringWithRange(Range<String.Index>( start:self.startIndex.advancedBy(from), end:self.startIndex.advancedBy(from+length))) } } 

Answer 14

对于Swift 2.0 ，就是这样的：

 var string1 = "www.stackoverflow.com" var index1 = string1.endIndex.advancedBy(-4) var substring1 = string1.substringToIndex(index1) 

Answer 15

Swift 2.0下面的代码是在XCode 7.2上testing的。 请参阅底部的附件截图

 import UIKit class ViewController: UIViewController { override func viewDidLoad() { super.viewDidLoad() var mainText = "http://stackoverflow.com" var range = Range(start: mainText.startIndex.advancedBy(7), end: mainText.startIndex.advancedBy(24)) var subText = mainText.substringWithRange(range) //OR Else use below for LAST INDEX range = Range(start: mainText.startIndex.advancedBy(7), end: mainText.endIndex) subText = mainText.substringWithRange(range) } } 

Answer 16

我修改了andrewz的post，使其与Swift 2.0兼容（也许Swift 3.0）。 在我看来，这个扩展是更容易理解和类似于其他语言 （如PHP） 可用 。

 extension String { func length() -> Int { return self.lengthOfBytesUsingEncoding(NSUTF16StringEncoding) } func substring(from:Int = 0, to:Int = -1) -> String { var nto=to if nto < 0 { nto = self.length() + nto } return self.substringWithRange(Range<String.Index>( start:self.startIndex.advancedBy(from), end:self.startIndex.advancedBy(nto+1))) } func substring(from:Int = 0, length:Int) -> String { return self.substringWithRange(Range<String.Index>( start:self.startIndex.advancedBy(from), end:self.startIndex.advancedBy(from+length))) } } 

Answer 17

这是一个在Swift中获取子string的简单方法

 import UIKit var str = "Hello, playground" var res = NSString(string: str) print(res.substring(from: 4)) print(res.substring(to: 10))