iOS:字符串。从绝对位置获取行号和列号，反之亦然

Question

假设我有一个包含混合的 CR、LF 和 CRLF 换行符的文本。

像这样:“\n\n Lorem\r Ipsum\n 是\r\n 只是\n 打印\r 和排版行业的\n 虚拟\r\n 文本。\n\n” 。

我正在将此文本加载到简单的文本编辑器 (NSTextView/UITextView)。 视觉换行符看起来一样；只是一个新行。

我可以在简单的文本编辑器中浏览文本、选择文本、剪切、复制、粘贴......

问题:如何从绝对字符位置获取行和列编号(即选择NSRange) ？还有，如何从已知的行和列数字中获取绝对字符位置？

谢谢!

---

更新 1:

• line 和 column 数字 - 简单表示光标位置。
• line 和 column 编号 - 具有基于One 的编号。
• 绝对 字符位置 - 具有基于零的编号。

当前解决方案的示例代码。它从绝对字符位置计算行和列号，反之亦然。但它不会重新计算文本更改的映射。

struct TextString {

   struct Cursor {
      let line: Int
      let column: Int
   }

   struct Mapping {
      let lineNumber: Int
      let lineLength: Int
      let absolutePosition: Int

      fileprivate var absoluteStart: Int {
         return absolutePosition - lineLength
      }
   }

   let string: String
   private (set) var mappings: [Mapping] = []

   init(string: String) {
      self.string = string
      mappings = setupMappings()
   }
}

extension TextString {

   func cursor(from position: Int) -> Cursor? {
      guard position > 0 else {
         return nil
      }
      guard let mapping = mappings.first(where: { $0.absolutePosition >= position && $0.absoluteStart <= position }) else {
         return nil
      }
      let result = Cursor(line: mapping.lineNumber, column: position - mapping.absoluteStart)
      return result
   }

   func position(from cursor: Cursor) -> Int? {
      guard let line = mappings.element(at: cursor.line - 1) else {
         return nil
      }
      guard line.lineLength >= cursor.column else {
         return nil
      }
      let result = line.absoluteStart + cursor.column
      return result
   }
}

extension TextString {

   private func setupMappings() -> [Mapping] {
      var mappings: [Mapping] = []
      var line = 1
      var previousAbsolutePosition = 0
      var delta = 0
      let scanner = Scanner(string: string)
      scanner.charactersToBeSkipped = nil
      while !scanner.isAtEnd {
         if scanner.scanUpToCharacters(from: .newlines) != nil {
            let charactersLocation = scanner.scanLocation - delta
            if let newLines = scanner.scanCharacters(from: .newlines) {
               for index in 0..<newLines.count {
                  let absolutePosition = charactersLocation + 1 + index // `+1` is newLine itself
                  mappings.append(Mapping(lineNumber: line, lineLength: absolutePosition - previousAbsolutePosition,
                                          absolutePosition: absolutePosition))
                  previousAbsolutePosition = absolutePosition
                  line += 1
               }
               delta = scanner.scanLocation - previousAbsolutePosition
            } else {
               // Only happens when we at last line withot newline.
               let absolutePosition = charactersLocation
               mappings.append(Mapping(lineNumber: line, lineLength: absolutePosition - previousAbsolutePosition,
                                       absolutePosition: absolutePosition))
               line += 1
               previousAbsolutePosition = charactersLocation
            }
         } else if let newLines = scanner.scanCharacters(from: .newlines) { // Text begins with new lines.
            for index in 0..<newLines.count {
               let absolutePosition = 1 + index // `+1` is newLine itself
               mappings.append(Mapping(lineNumber: line, lineLength: absolutePosition - previousAbsolutePosition,
                                       absolutePosition: absolutePosition))
               previousAbsolutePosition = absolutePosition
               line += 1
            }
            delta = scanner.scanLocation - previousAbsolutePosition
         }
      }
      assert(previousAbsolutePosition == string.count)
      return mappings
   }
}

---

更新 2:RegEx 版本。

private func setupMappingsUsingRegex() throws -> [Mapping] {
   if string.isEmpty {
      return []
   }
   var mappings: [Mapping] = []
   let regex = try NSRegularExpression(pattern: "(\\r\\n)|(\\n)|(\\r)")
   let matches = regex.matches(in: string, range: NSRange(location: 0, length: string.unicodeScalars.count))
   var line = 1
   var previousAbsolutePosition = 0
   var delta = 0

   // String without any newline.
   if matches.isEmpty {
      let mapping = Mapping(lineNumber: 1, lineLength: string.count, absolutePosition: string.count)
      mappings.append(mapping)
      return mappings
   }

   for match in matches {
      let absolutePosition = match.range.location - delta + 1
      let mapping = Mapping(lineNumber: line, lineLength: absolutePosition - previousAbsolutePosition,
                            absolutePosition: absolutePosition)
      mappings.append(mapping)
      delta += match.range.length - 1
      previousAbsolutePosition = absolutePosition
      line += 1
   }

   // Rest of the string without newline at the end.
   if previousAbsolutePosition < string.count {
      let mapping = Mapping(lineNumber: line, lineLength: string.count - previousAbsolutePosition,
                            absolutePosition: string.count)
      mappings.append(mapping)
      previousAbsolutePosition = string.count
   }
   assert(previousAbsolutePosition == string.count)
   return mappings
}

---

性能:22400 个字符(200 行)分析 1000 次。

• 正则表达式:5.120s
• 扫描器:6.603s

Answer 1

我建议您使用正则表达式分隔字符串。如果您看到 \n、\r 和 \r\n，假设您想拆分一个子字符串，则正则表达式类似于

var content: String = <Your text here>
let regex = try! NSRegularExpression(pattern: "(\\n)|(\\r)|(\\r\\n)")
let matchs = regex.matches(in: content, range: NSRange(location: 0, length: content.count)).map{(content as NSString).substring(with: $0.range)}

然后你可以在匹配的结果中循环并获取索引和范围等