Question

我想要编辑以下字符串：

var someString = "I wan't this text {something I don't want}"

我想删除两个大括号中包含的所有文本，无论文本有多长。当我知道范围时，我一直在使用以下代码删除字符串的一部分：

extension String {
    mutating func deleteCharactersInRange(range: NSRange) {
        let mutableSelf = NSMutableString(string: self)
        mutableSelf.deleteCharactersInRange(range)
        self = mutableSelf
    }
}

但是，我不知道问题的范围。有什么想法吗？

Answer 1

将NSString和NSRange与Swift的String和Range混合使用时，使用字符串和范围可能非常具有挑战性。

这是一个纯粹的Swift解决方案。

var someString = "I wan't this text {something I don't want}"

let rangeOpenCurl = someString.rangeOfString("{")
let rangeCloseCurl = someString.rangeOfString("}")
if let startLocation = rangeOpenCurl?.startIndex,
    let endLocation = rangeCloseCurl?.endIndex {
    someString.replaceRange(startLocation ..< endLocation, with: "")
}

Answer 2

使用RegEx模式匹配用大括号括起的任何内容：

var sourceString: String = "I wan\'t this text {something I don't want}"

let destinationString = sourceString.stringByReplacingOccurrencesOfString("\\{(.*?)\\}", withString: "", options: .RegularExpressionSearch)

print(destinationString)

这将打印“我不是这个文字”，没有双引号。

Answer 3

extension String {
    func getCurlyBraceRanges() -> [NSRange] {
        var results = [NSRange]()
        var leftCurlyBrace = -1
        for index in 0..<self.characters.count {
            let char = self[self.startIndex.advancedBy(index)]
            if char == Character("{") {
                leftCurlyBrace = index
            } else if char == Character("}") {
                if leftCurlyBrace != -1 {
                    results.append(NSRange(location: leftCurlyBrace, length: index - leftCurlyBrace + 1))
                    leftCurlyBrace = -1
                }
            }

        }
        return results
    }
    mutating func deleteCharactersInRange(range: NSRange) {
        let mutableSelf = NSMutableString(string: self)
        mutableSelf.deleteCharactersInRange(range)
        self = String(mutableSelf)
    }
    mutating func deleteCharactersInRanges(ranges: [NSRange]) {
        var tmpString = self
        for i in (0..<ranges.count).reverse() {
            tmpString.deleteCharactersInRange(ranges[i])
            print(tmpString)
        }
        self = tmpString
    }
}

var testString = "I wan't this text {something I don't want}"

testString.deleteCharactersInRanges(testString.getCurlyBraceRanges())

输出：＆＃34;我想要这个文字＆＃34;

删除字符串范围内的字符

3 个答案: