【发布时间】：2019-03-01 17:02:04
【问题描述】：
假设我有一个包含混合 CR、LF 和 CRLF 换行符分隔符的文本。
像这样："\n \n Lorem \r Ipsum \n is \r\n simply \n dummy \r\n text of \n the printing \r and typesetting industry. \n \n"。
我正在将此文本加载到简单的文本编辑器 (NSTextView/UITextView)。
视觉上换行符看起来是一样的；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absolute 字符位置（即选择NSRange）获取line 和column 编号？另外，如何从已知的line 和column 号码中获取absolute 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 - 具有基于一个的编号。 -
absolute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absolute 字符位置计算line 和column 数字，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：5.120 秒
- 扫描仪：6.603 秒
【问题讨论】：
-
我会说使用stackoverflow.com/questions/31746223/…，计算
\r\n（首先）的数量，然后是\n和`r"，您应该能够从由“0 到 absoluteLocation”组成的子字符串，与迭代获得 absoluteLocation 的方式相同？我不确定“行”和“列”的定义，但您可能想举个例子，比如说“s”simply. -
混合行结尾是没有意义的。我相信文本视图只识别\n。您应该将行尾规范化为 \n ，然后解决您的问题。