如何在 Swift 4 中使用 String 子串？ 'substring(to:)' 已弃用：请使用带有 'partial range from' 运算符的字符串切片下标

Question

我在 Swift 3 中编写了以下简单代码：

let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)

从 Xcode 9 beta 5 开始，我收到以下警告：

'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator.

如何在 Swift 4 中使用 部分范围为 的切片下标？

Answer 1

你应该留空一侧，因此得名"partial range"。

let newStr = str[..<index]

同样代表partial range from operators, 只需将另一边留空即可：

let newStr = str[index...]

请记住，这些范围运算符 return 和 Substring。如果要转成字符串，使用String的初始化函数：

let newStr = String(str[..<index])

您可以阅读有关新子字符串的更多信息 here。

Answer 2

Swift3 和 Swift4 中 uppercasedFirstCharacter 便利属性的示例。

属性 uppercasedFirstCharacterNew 演示如何在Swift4中使用String切片下标

extension String {

   public var uppercasedFirstCharacterOld: String {
      if characters.count > 0 {
         let splitIndex = index(after: startIndex)
         let firstCharacter = substring(to: splitIndex).uppercased()
         let sentence = substring(from: splitIndex)
         return firstCharacter + sentence
      } else {
         return self
      }
   }

   public var uppercasedFirstCharacterNew: String {
      if characters.count > 0 {
         let splitIndex = index(after: startIndex)
         let firstCharacter = self[..<splitIndex].uppercased()
         let sentence = self[splitIndex...]
         return firstCharacter + sentence
      } else {
         return self
      }
   }
}

let lorem = "lorem".uppercasedFirstCharacterOld
print(lorem) // Prints "Lorem"

let ipsum = "ipsum".uppercasedFirstCharacterNew
print(ipsum) // Prints "Ipsum"

Answer 3

将您的代码转换为 Swift 4 也可以这样完成：

let str = "Hello, playground"
let index = str.index(of: ",")!
let substr = str.prefix(upTo: index)

您可以使用下面的代码来获得一个新字符串：

let newString = String(str.prefix(upTo: index))

Answer 4

substring(from: index) 转换为 [index...]

检查样本

let text = "1234567890"
let index = text.index(text.startIndex, offsetBy: 3)

text.substring(from: index) // "4567890"   [Swift 3]
String(text[index...])      // "4567890"   [Swift 4]

Answer 5

一些有用的扩展：

extension String {
    func substring(from: Int, to: Int) -> String {
        let start = index(startIndex, offsetBy: from)
        let end = index(start, offsetBy: to - from)
        return String(self[start ..< end])
    }

    func substring(range: NSRange) -> String {
        return substring(from: range.lowerBound, to: range.upperBound)
    }
}

Answer 6

将子字符串 (Swift 3) 转换为字符串切片 (Swift 4)

Swift3、4 中的示例：

let newStr = str.substring(to: index) // Swift 3
let newStr = String(str[..<index]) // Swift 4

let newStr = str.substring(from: index) // Swift 3
let newStr = String(str[index...]) // Swift 4

let range = firstIndex..<secondIndex // If you have a range
let newStr = = str.substring(with: range) // Swift 3
let newStr = String(str[range])  // Swift 4

Answer 7

Swift 5, 4

用法

let text = "Hello world"
text[0] // H
text[...3] // "Hell"
text[6..<text.count] // world
text[NSRange(location: 6, length: 3)] // wor

代码

import Foundation

public extension String {
  subscript(value: Int) -> Character {
    self[index(at: value)]
  }
}

public extension String {
  subscript(value: NSRange) -> Substring {
    self[value.lowerBound..<value.upperBound]
  }
}

public extension String {
  subscript(value: CountableClosedRange<Int>) -> Substring {
    self[index(at: value.lowerBound)...index(at: value.upperBound)]
  }

  subscript(value: CountableRange<Int>) -> Substring {
    self[index(at: value.lowerBound)..<index(at: value.upperBound)]
  }

  subscript(value: PartialRangeUpTo<Int>) -> Substring {
    self[..<index(at: value.upperBound)]
  }

  subscript(value: PartialRangeThrough<Int>) -> Substring {
    self[...index(at: value.upperBound)]
  }

  subscript(value: PartialRangeFrom<Int>) -> Substring {
    self[index(at: value.lowerBound)...]
  }
}

private extension String {
  func index(at offset: Int) -> String.Index {
    index(startIndex, offsetBy: offset)
  }
}

Answer 8

在编程时，我经常使用纯 A-Za-z 和 0-9 的字符串。不需要困难的索引操作。此扩展基于普通的旧左/中/右功能。

extension String {

    // LEFT
    // Returns the specified number of chars from the left of the string
    // let str = "Hello"
    // print(str.left(3))         // Hel
    func left(_ to: Int) -> String {
        return "\(self[..<self.index(startIndex, offsetBy: to)])"
    }

    // RIGHT
    // Returns the specified number of chars from the right of the string
    // let str = "Hello"
    // print(str.left(3))         // llo
    func right(_ from: Int) -> String {
        return "\(self[self.index(startIndex, offsetBy: self.length-from)...])"
    }

    // MID
    // Returns the specified number of chars from the startpoint of the string
    // let str = "Hello"
    // print(str.left(2,amount: 2))         // ll
    func mid(_ from: Int, amount: Int) -> String {
        let x = "\(self[self.index(startIndex, offsetBy: from)...])"
        return x.left(amount)
    }
}

Answer 9

这是我的解决方案，没有警告，没有错误，但是很完美

let redStr: String = String(trimmStr[String.Index.init(encodedOffset: 0)..<String.Index.init(encodedOffset: 2)])
let greenStr: String = String(trimmStr[String.Index.init(encodedOffset: 3)..<String.Index.init(encodedOffset: 4)])
let blueStr: String = String(trimmStr[String.Index.init(encodedOffset: 5)..<String.Index.init(encodedOffset: 6)])

Answer 10

我写了一个字符串扩展来替换'String: subString:'

extension String {
    
    func sliceByCharacter(from: Character, to: Character) -> String? {
        let fromIndex = self.index(self.index(of: from)!, offsetBy: 1)
        let toIndex = self.index(self.index(of: to)!, offsetBy: -1)
        return String(self[fromIndex...toIndex])
    }
    
    func sliceByString(from:String, to:String) -> String? {
        //From - startIndex
        var range = self.range(of: from)
        let subString = String(self[range!.upperBound...])
        
        //To - endIndex
        range = subString.range(of: to)
        return String(subString[..<range!.lowerBound])
    }
    
}

Usage : "Date(1511508780012+0530)".sliceByString(from: "(", to: "+")

Example Result : "1511508780012"

PS：可选项被强制解包。请在必要时添加类型安全检查。

Answer 11

使用 Swift 4:

从字符串创建子字符串（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 :

如果要在 2 个索引之间生成子字符串，请使用：

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

Answer 12

您可以使用 class 字符串的扩展来创建自定义子字符串方法，如下所示：

extension String {
    func subString(startIndex: Int, endIndex: Int) -> String {
        let end = (endIndex - self.count) + 1
        let indexStartOfText = self.index(self.startIndex, offsetBy: startIndex)
        let indexEndOfText = self.index(self.endIndex, offsetBy: end)
        let substring = self[indexStartOfText..<indexEndOfText]
        return String(substring)
    }
}

Answer 13

希望这会有所帮助:-

var string = "123456789"

如果你想要在某个特定索引之后的子字符串。

var indexStart  =  string.index(after: string.startIndex )// you can use any index in place of startIndex
var strIndexStart   = String (string[indexStart...])//23456789

如果你想在最后删除一些字符串后得到一个子字符串。

var indexEnd  =  string.index(before: string.endIndex)
var strIndexEnd   = String (string[..<indexEnd])//12345678

您还可以使用以下代码创建索引:-

var  indexWithOffset =  string.index(string.startIndex, offsetBy: 4)

Answer 14

Swift 4/5 更短：

let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"

Answer 15

使用此方法，您可以获得特定范围的 string.you 需要传递起始索引，然后是您想要的字符总数。

extension String{
    func substring(fromIndex : Int,count : Int) -> String{
        let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
        let endIndex = self.index(self.startIndex, offsetBy: fromIndex + count)
        let range = startIndex..<endIndex
        return String(self[range])
    }
}

Answer 16

希望对您有所帮助。

extension String {
    func getSubString(_ char: Character) -> String {
        var subString = ""
        for eachChar in self {
            if eachChar == char {
                return subString
            } else {
                subString += String(eachChar)
            }
        }
        return subString
    }
}


let str: String = "Hello, playground"
print(str.getSubString(","))

Answer 17

斯威夫特5

(Java的子串方法):

extension String {
    func subString(from: Int, to: Int) -> String {
       let startIndex = self.index(self.startIndex, offsetBy: from)
       let endIndex = self.index(self.startIndex, offsetBy: to)
       return String(self[startIndex..<endIndex])
    }
}

用法：

var str = "Hello, Nick Michaels"
print(str.subString(from:7,to:20))
// print Nick Michaels

Answer 18

var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))

你可以这样试一下，会得到好的结果。

Answer 19

如果你只想得到一个特定字符的子串，你不需要先找到索引，你可以使用prefix(while:)方法

let str = "Hello, playground"
let subString = str.prefix { [=10=] != "," } // "Hello" as a String.SubSequence

Answer 20

我使用的最简单的方法是：

String(Array(str)[2...4])

如何在 Swift 4 中使用 String 子串？ 'substring(to:)' 已弃用：请使用带有 'partial range from' 运算符的字符串切片下标

How can I use String substring in Swift 4? 'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator

swift

swift4

Swift 5, 4

用法

代码

使用 Swift 4:

输出

如果要在 2 个索引之间生成子字符串，请使用：

斯威夫特5

(Java的子串方法):