Swift 4个字符范围

Question

我正在尝试在字符串中制作一个函数，从而从提供的字符串中删除某些字符。

例如，我有

"‌Boll %b‌ (Teeth ‌Alligator‌ (13,8,8,5,5,3,n),20,2,ma)"

我想编辑

"‌Boll %b‌"

这意味着要删除"("和")"

之间的所有内容

所以我已经完成了此功能

func deleteInnerString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString
}

但是在这种情况下，输出字符串为

"‌Boll %b‌ ,20,2,ma)"

问题是，如何扫描字符串以找到最后一个闭合括号范围??

Answer 1

请注意，您的策略与非嵌套括号无法使用。

例如。此输入

a(b)c(d)e

使用您的方法将产生此输出

ae

哪个恕我直言不正确。

解决方案

此解决方案将与嵌套和非嵌套 parenthesis

一起使用。

let text = "‌a(b)c(d)e"
var numOfNestedParentesis = 0
var indexesToRemove:Set<Int> = []
for (index, char) in text.enumerated() {
    if char == "(" {
        numOfNestedParentesis += 1
    }
    if numOfNestedParentesis > 0 {
        indexesToRemove.insert(index)
    }
    if char == ")" {
        numOfNestedParentesis -= 1
    }
}
let result = String(text.enumerated()
    .filter { !indexesToRemove.contains($0.offset) }
    .map { $0.element } )

测试

print(result) // ‌ace

Answer 2

String.range功能可以提供额外的选项，因此您可以将其传递给String.CompareOptions.backwards参数以获取最后一个实例。

func deleteInnerString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString
}

Answer 3

您必须在字符串的 range函数中使用 .backwards的 CC_5选项。如下

func deleteString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString.trimmingCharacters(in: .whitespaces)
}