这在Swift 5中为我工作。用例是从剪贴板或类似的地方获取URL，该URL可能已经有转义字符，但还包含Unicode字符，这可能导致URLComponents或URL(string:)失败。

首先，创建一个包含所有url合法字符的字符集:

extension CharacterSet {

    /// Characters valid in at least one part of a URL.
    ///
    /// These characters are not allowed in ALL parts of a URL; each part has different requirements. This set is useful for checking for Unicode characters that need to be percent encoded before performing a validity check on individual URL components.
    static var urlAllowedCharacters: CharacterSet {
        // Start by including hash, which isn't in any set
        var characters = CharacterSet(charactersIn: "#")
        // All URL-legal characters
        characters.formUnion(.urlUserAllowed)
        characters.formUnion(.urlPasswordAllowed)
        characters.formUnion(.urlHostAllowed)
        characters.formUnion(.urlPathAllowed)
        characters.formUnion(.urlQueryAllowed)
        characters.formUnion(.urlFragmentAllowed)

        return characters
    }
}

接下来，用一个方法扩展String来编码url:

extension String {

    /// Converts a string to a percent-encoded URL, including Unicode characters.
    ///
    /// - Returns: An encoded URL if all steps succeed, otherwise nil.
    func encodedUrl() -> URL? {        
        // Remove preexisting encoding,
        guard let decodedString = self.removingPercentEncoding,
            // encode any Unicode characters so URLComponents doesn't choke,
            let unicodeEncodedString = decodedString.addingPercentEncoding(withAllowedCharacters: .urlAllowedCharacters),
            // break into components to use proper encoding for each part,
            let components = URLComponents(string: unicodeEncodedString),
            // and reencode, to revert decoding while encoding missed characters.
            let percentEncodedUrl = components.url else {
            // Encoding failed
            return nil
        }

        return percentEncodedUrl
    }

}

可以这样测试:

let urlText = "https://www.example.com/폴더/search?q=123&foo=bar&multi=eggs+and+ham&hangul=한글&spaced=lovely%20spam&illegal=<>#top"
let url = encodedUrl(from: urlText)

url最后的值:https://www.example.com/%ED%8F%B4%EB%8D%94/search?q=123&foo=bar&multi=eggs+and+ham&hangul=%ED%95%9C%EA%B8%80&spaced=lovely%20spam&illegal=%3C%3E#top

注意，%20和+空格都被保留，Unicode字符被编码，原始urlText中的%20没有被双重编码，锚(片段或#)仍然保留。

编辑:现在检查每个组件的有效性。

你可以使用URLComponents来避免手动对查询字符串进行百分比编码:

let scheme = "https"
let host = "www.google.com"
let path = "/search"
let queryItem = URLQueryItem(name: "q", value: "Formula One")


var urlComponents = URLComponents()
urlComponents.scheme = scheme
urlComponents.host = host
urlComponents.path = path
urlComponents.queryItems = [queryItem]

if let url = urlComponents.url {
    print(url)   // "https://www.google.com/search?q=Formula%20One"
}

extension URLComponents {
    init(scheme: String = "https",
         host: String = "www.google.com",
         path: String = "/search",
         queryItems: [URLQueryItem]) {
        self.init()
        self.scheme = scheme
        self.host = host
        self.path = path
        self.queryItems = queryItems
    }
}

let query = "Formula One"
if let url = URLComponents(queryItems: [URLQueryItem(name: "q", value: query)]).url {
    print(url)  // https://www.google.com/search?q=Formula%20One
}

Swift 4 & 5

要在URL中编码参数，我发现使用. alphanumics字符集是最简单的选择:

let urlEncoded = value.addingPercentEncoding(withAllowedCharacters: .alphanumerics)
let url = "http://www.example.com/?name=\(urlEncoded!)"

使用URL编码的任何标准字符集(如. urlqueryallowed或. urlhostallowed)都不起作用，因为它们不排除=或&字符。

注意，通过使用. alphannumeric，它将编码一些不需要编码的字符(如-，.，_或~ -参见2.3。RFC 3986中的非保留字符)。我发现使用. alphanumics比构造一个自定义字符集更简单，并且不介意编码一些额外的字符。如果这困扰你，构造一个自定义字符集，如描述如何百分比编码URL字符串，例如:

// Store allowed character set for reuse (computed lazily).
private let urlAllowed: CharacterSet =
    .alphanumerics.union(.init(charactersIn: "-._~")) // as per RFC 3986

extension String {
    var urlEncoded: String? {
        return addingPercentEncoding(withAllowedCharacters: urlAllowed)
    }
}

let url = "http://www.example.com/?name=\(value.urlEncoded!)"

警告:urlEncoded参数是强制打开的。对于无效的unicode字符串，它可能会崩溃。参见为什么String.addingPercentEncoding()的返回值是可选的?而不是强制展开urlEncoded!你可以使用urlEncoded ??或者if let urlEncoded = urlEncoded{…}。

版本:斯威夫特5

// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:@&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring  

斯威夫特5:

extension String {
    var urlEncoded: String? {
        let allowedCharacterSet = CharacterSet.alphanumerics.union(CharacterSet(charactersIn: "~-_."))
        return self.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet)
    }
}

print("\u{48}ello\u{9}world\u{7}\u{0}".urlEncoded!) // prints Hello%09world%07%00
print("The string ü@foo-bar".urlEncoded!) // prints The%20string%20%C3%BC%40foo-bar

帮助我的是，我创建了一个单独的NSCharacterSet，并在一个UTF-8编码的字符串上使用它，即textToEncode来生成所需的结果:

var queryCharSet = NSCharacterSet.urlQueryAllowed
queryCharSet.remove(charactersIn: "+&?,:;@+=$*()")
    
let utfedCharacterSet = String(utf8String: textToEncode.cString(using: .utf8)!)!
let encodedStr = utfedCharacterSet.addingPercentEncoding(withAllowedCharacters: queryCharSet)!
    
let paramUrl = "https://api.abc.eu/api/search?device=true&query=\(escapedStr)"