Swift 3.1 获取两个字符之间的范围？

Question

我显然做错了什么，因为这让我花了很长时间来做以下事情:(下面的 Playground 代码)。

注意:Swift 3.1 ~

我只是想从 piz(123)zazz 中取出 123

let aString = "piz(123)zazz"

let startBracket: Character = "("
if let idx1 = aString.characters.index(of: startBracket) {
    let pos1 = aString.characters.distance(from: aString.startIndex, to: idx1)
    print("Found \(startBracket) at position \(pos1)")
}
else {
    print("Not found")
}

let endBracket: Character = ")"
if let idx2 = aString.characters.index(of: endBracket) {
    let pos2 = aString.characters.distance(from: aString.startIndex, to: idx2)
    print("Found \(startBracket) at position \(pos2)")
}
else {
    print("Not found")
}

let range = pos1..<pos2 // << this is not working, I give up!!!

let result_1 = aString.substring(with: range)

Answer 1

这更容易:

• 起始索引为字符串中(范围
的 upperBound

• 结束索引为字符串中)范围的lowerBound

  let aString = "piz(123)zazz"
  
  if let openParenthesisRange = aString.range(of: "("),
      let closeParenthesisRange = aString.range(of: ")", range: openParenthesisRange.upperBound..<aString.endIndex) {
  
      let range = openParenthesisRange.upperBound..<closeParenthesisRange.lowerBound
      let result = aString.substring(with: range)
      print(result)
  }
  else {
      print("Not found")
  }
  

---

替代正则表达式，它的代码更多但用途更广

let string = "piz(123)zazz"

let pattern = "\\((\\d+)\\)" // searches for 0 ore more digits between parentheses

do {
    let regex = try NSRegularExpression(pattern: pattern)
    if let match = regex.firstMatch(in: string, range: NSRange(location: 0, length: string.utf16.count)) {
        let range = match.rangeAt(1)
        let start = string.index(string.startIndex, offsetBy: range.location)
        let end = string.index(start, offsetBy: range.length)
        print(string.substring(with: start..<end))
    } else {
        print("Not Found")
    }
} catch {
    print("Regex Error:", error)
}