这就简单多了：

• 开始索引是字符串中(范围的upperBound
• 结束索引是字符串中)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")
  }

字符串
或者正则表达式，它的代码更多，但它更通用

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(string.startIndex..., in: string)) {
        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)
}

型

当您尝试使用时，您的pos1和pos2超出了作用域。它们都已在if块内定义，并且它们的作用域在此结束。
我建议在顶部声明它们，以便将它们纳入范围。

let aString = "piz(123)zazz"
 let pos1: Int?  // You don't need optional here but I prefer it 
 let pos2: Int? 
 .
 .
 .
 if let p1 = pos1, p2 = pos2  { // Optional Chaining
     let range = pos1..<pos2
 }

字符串