将UTF-8(字节)emoji代码转换为文本形式的emoji图标

因为你已经给出了转换器工具的链接 UTF-8 编码和解码。你有 UTF-8 编码字符串,因此这里是 UTF8-Decoding .

目标-C

const char *ch = [@"Hii \xF0\x9F\x98\x81" cStringUsingEncoding:NSUTF8StringEncoding];
NSString *decode_string = [NSString stringWithUTF8String:ch];
NSLog(@"%@",decode_string);

输出: 小精灵

迅捷

我可以转换 \\xF0\\x9F\\x98\\x81 到 😁 在里面 SWift . 首先我把六弦转换成 Data 然后回到 String 使用 UTF-8 编码。

var str = "\\xF0\\x9F\\x98\\x81"
if let data = data(fromHexaStr: str) {
     print(String(data: data, encoding: String.Encoding.utf8) ?? "")
}

输出: __

下面是我用来将十六进制字符串转换为数据的函数。我跟着去了 this answer .

func data(fromHexaStr hexaStr: String) -> Data? {
    var data = Data(capacity: hexaStr.characters.count / 2)
    let regex = try! NSRegularExpression(pattern: "[0-9a-f]{1,2}", options: .caseInsensitive)
    regex.enumerateMatches(in: hexaStr, range: NSMakeRange(0, hexaStr.utf16.count)) { match, flags, stop in
        let byteString = (hexaStr as NSString).substring(with: match!.range)
        var num = UInt8(byteString, radix: 16)!
        data.append(&num, count: 1)
    }

    guard data.count > 0 else { return nil }

    return data
}

注: 上面的代码的问题是它只转换十六进制字符串而不是组合字符串。

最终工作方案:Swift

我用for循环代替了 [0-9a-f]{1,2} 因为这也会扫描 81, 9F, Any Two digits number 这显然是错误的。

例如: I have 81 INR \\xF0\\x9F\\x98\\x81 .

/// This line will convert "F0" into hexa bytes
let byte = UInt8("F0", radix: 16)

我做了一个字符串扩展,如果它有前缀,我每4个字符检查一次 \x 和计数 4 最后两个字符通过使用 radix 如上所述。

extension String {

    func hexaDecoededString() -> String {

        var newData = Data()
        var emojiStr: String = ""
        for char in self.characters {

            let str = String(char)
            if str == "\\" || str.lowercased() == "x" {
                emojiStr.append(str)
            }
            else if emojiStr.hasPrefix("\\x") || emojiStr.hasPrefix("\\X") {
                emojiStr.append(str)
                if emojiStr.count == 4 {
                    /// It can be a hexa value
                    let value = emojiStr.replacingOccurrences(of: "\\x", with: "")
                    if let byte = UInt8(value, radix: 16) {
                        newData.append(byte)
                    }
                    else {
                        newData.append(emojiStr.data(using: .utf8)!)
                    }
                    /// Reset emojiStr
                    emojiStr = ""
                }
            }
            else {
                /// Append the data as it is
                newData.append(str.data(using: .utf8)!)
            }
        }

        let decodedString = String(data: newData, encoding: String.Encoding.utf8)
        return decodedString ?? ""
    }
}

用途:

var hexaStr = "Hi \\xF0\\x9F\\x98\\x81 81"
print(hexaStr.hexaDecoededString())

嗨81

hexaStr = "Welcome to SP19!\\xF0\\x9f\\x98\\x81"
print(hexaStr.hexaDecoededString())

欢迎来到SP19!

我解决了你的问题,但需要更多的工作才能使它成为一般性的问题,这里的问题是你的emijo代表了 Hex Byte x9F ,所以我们必须将这个十六进制转换为 utf8 然后将其转换为 Data 最后将数据转换为 String

最终结果 Hii 😁 请阅读评论

 let strTemp = "Hii \\xF0\\x9F\\x98\\x81"


            let regex = try! NSRegularExpression(pattern: "[0-9a-f]{1,2}", options: .caseInsensitive)
            // get all matched hex  xF0 , x9f,..etc

            let matches = regex.matches(in: strTemp, options: [], range: NSMakeRange(0, strTemp.count))


            // Data that will hanlde convert hex to UTf8
            var emijoData = Data(capacity: strTemp.count / 2)

            matches.enumerated().forEach { (offset , check) in
                let byteString = (strTemp as NSString).substring(with: check.range)
                var num = UInt8(byteString, radix: 16)!
                emijoData.append(&num, count: 1)
            }

            let subStringEmijo = String.init(data: emijoData, encoding: String.Encoding.utf8)!
            //now we have your emijo text  😁 we can replace by its code from string using matched ranges `first` and `last`

            // All range range of  \\xF0\\x9F\\x98\\x81 in "Hii \\xF0\\x9F\\x98\\x81" to replce by your emijo

            if let start = matches.first?.range.location, let end = matches.last?.range.location  , let endLength = matches.last?.range.length {

                let startLocation = start  - 2
                let length = end - startLocation + endLength

                let sub = (strTemp as NSString).substring(with: NSRange.init(location: startLocation, length: length))

                print( strTemp.replacingOccurrences(of: sub, with: subStringEmijo))
              // Hii 😁

            }