Swift替换字符串中的第一个字符

7 回答

• 12

  你可以试试这个：

  var s = "123456"
  
  let s2 = s.replacingCharacters(in: ...s.startIndex, with: "a")
  
  s2 // "a23456"
  

  回复于 2024-04-16T14:53:58+08:00
• 16

  Swift 3

  let string = "&whatever"
  let output= "?" + String(string.characters.dropFirst())
  

  Swift 4 or later

  let string = "&whatever"
  let output = "?" + string.dropFirst()
  

  回复于 2024-04-16T14:53:58+08:00
• 0

  let range = Range(start: query.startIndex, end: query.startIndex.successor())
  query.replaceRange(range, with: "?")
  

  或更短

  let s = query.startIndex
  query.replaceRange(s...s, with: "?")
  

  这与Objective-C有点不同 - 但我们会习惯它;-) .

  如果您对此感到不舒服，可以随时回退到旧的 NSString API：

  let newQuery = (query as NSString).stringByReplacingCharactersInRange(
    NSMakeRange(0,1), withString: "?") // as String not necessary
  

  回复于 2024-04-16T14:53:58+08:00
• 1

  在Swift 4.0中

  let inputString = "!Hello"
  let outputString = String(inputString.dropFirst())
  

  Answer

  outputString = "Hello"
  

  回复于 2024-04-16T14:53:58+08:00
• 4

  问题是您传递的是NSRange值，其中需要Range对象 . 使用Swift字符串时，必须使用

  Range<String.Index>(start: str.startIndex, end: advance(str.startIndex, 1))
  

  回复于 2024-04-16T14:53:58+08:00
• 0

  在Swift 2.0中你可以这样做： -

  let inputString = "!Hello"
  let outputString = String(inputString.characters.dropFirst())
  
  //Answer
  outputString = "Hello"
  

  回复于 2024-04-16T14:53:58+08:00
• 0

  Swift 4.0

  extension String
  {
      func replace_fromStart(str:String , endIndex:Int , With:String) -> String {
          var strReplaced = str ;
          let start = str.startIndex;
          let end = str.index(str.startIndex, offsetBy: endIndex);
          strReplaced = str.replacingCharacters(in: start..<end, with: With) ;
          return strReplaced;
      }
  }
  

  在任何地方使用它

  MY_String.replace_fromStart(str: MY_String, endIndex: 10, With: "**********")
  

  回复于 2024-04-16T14:53:58+08:00