如何检查字符串中的最后3个字符是否包含Swift中的数字？

5 回答

• 1

  （正如@NiravD指出的，用于预夫特3时，使用 where 加入的多子句的条件部分 . 用于SWIFT 3，多条条件部分被简单地通过加入 , . 对于以下两种方法中，都夫特2.2和3版本包括在内）

  ---

  对数字字符使用模式匹配“0”...“9”

  Swift 2.2

  extension String {
      var lastThreeLettersAreNumbers: Bool {
          if case let chars = characters.suffix(3) where chars.count > 2 {
              let numbersPattern = Character("0")..."9"
              return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
          }
          return false
      }
  }
  

  Swift 3

  extension String {
      var lastThreeLettersAreNumbers: Bool {
          if case let chars = characters.suffix(3), chars.count > 2 {
              let numbersPattern = Character("0")..."9"
              return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
          }
          return false
      }
  }
  
  /* example usage, common for both Swift 2.2/3 version */
  let str1 = "SACH092"
  let str2 = "SACH0B2"
  
  print(str1.lastThreeLettersAreNumbers) // true
  print(str2.lastThreeLettersAreNumbers) // false
  

  ---

  使用flat初始化器使用nil-return Int和flatMap

  您可以利用 Int by String 初始化程序为无法表示为整数的字符串返回 nil 这一事实 .

  Swift 2.2

  extension String {
      var lastThreeLettersAreNumbers: Bool {
          if case let chars = characters.suffix(3) where chars.count > 2 {
              return chars.flatMap{Int(String($0))}.count == 3
          }
          return false
      }
  }
  

  Swift 3

  extension String {
      var lastThreeLettersAreNumbers: Bool {
          if case let chars = characters.suffix(3), chars.count > 2 {
              return chars.flatMap{Int(String($0))}.count == 3
          }
          return false
      }
  }
  
  /* example usage, common for both Swift 2.2/3 version */
  let str1 = "SACH092"
  let str2 = "SACH0B2"
  
  print(str1.lastThreeLettersAreNumbers) // true
  print(str2.lastThreeLettersAreNumbers) // false
  

  回复于 2024-04-26T11:46:57+08:00
• 3

  要获得最后3个字符，

  let exampleString = "SACH092"
  let last3Char = exampleString.substringFromIndex(exampleString.endIndex.advancedBy(-3))
  

  检查last3Char是否包含所有数字，

  let badCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
  
  if last3Char.rangeOfCharacterFromSet(badCharacters) == nil {
      print("String contains all digits")
  } else {
      print("String contains non-digit characters")
  }
  

  回复于 2024-04-26T11:46:57+08:00
• 0

  实现字符串控制的最佳方法是使用正则表达式 .

  为了你 ：

  var str = "SACH092"
  let pattern = "^.*[0-9]{3,3}$"
  let regexp = try! NSRegularExpression(pattern: pattern, options: [])
  let matches = regexp.matches(in: str, options: [], range: NSMakeRange(0, str.characters.count))
  print("End with 3 numbers : \(matches.count > 0)")
  

  回复于 2024-04-26T11:46:57+08:00
• 1

  你可以像这样使用

  let s : NSString = "SACH092"
  let trimmedString: String = (s as NSString).substringFromIndex(max(s.length-3,0))
      print(trimmedString.isNumeric) // return true or false
  

  //扩展字符串

  extension String {
      var isNumeric: Bool {
          let nums: Set<Character> = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
          return Set(self.characters).isSubsetOf(nums)
      }
  }
  

  回复于 2024-04-26T11:46:57+08:00
• 0

  获取字符串的最后三个字母

  let oldString = "yourString"
  let newString = a.substringFromIndex(a.endIndex.advancedBy(-3))
  

  要检查这些字符是数字，

  func isNumber(num: String) -> Bool {
      let numberCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
      return !num.isEmpty && num.rangeOfCharacterFromSet(numberCharacters) == nil
  }
  

  回复于 2024-04-26T11:46:57+08:00