斯威夫特4.2

有时发生这种情况只是因为在段码中有空间或没有URL编码的参数通过API URL。

let myString = self.slugValue
                let csCopy = CharacterSet(bitmapRepresentation: CharacterSet.urlPathAllowed.bitmapRepresentation)
                let escapedString = myString!.addingPercentEncoding(withAllowedCharacters: csCopy)!
                //always "info:hello%20world"
                print(escapedString)

注意:不要忘记探索位图表示。

这个对我有用。

func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {

    let unreserved = "*-._"
    let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
    allowed.addCharactersInString(unreserved)

    if plusForSpace {
        allowed.addCharactersInString(" ")
    }

    var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)

    if plusForSpace {
        encoded = encoded?.stringByReplacingOccurrencesOfString(" ", withString: "+")
    }
    return encoded
}

我发现上述功能从这个链接:http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/。

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。

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  

让Url = Url。加德林编。“”)

斯威夫特4:

这取决于您的服务器所遵循的编码规则。

苹果提供了这个类方法，但它没有报告它遵循哪种RCF协议。

var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!

使用这个有用的工具，你应该保证这些字符的编码为你的参数:

$(美元符号)变为%24 &(&)变成%26 +(+)变成%2B ，(逗号)变成%2C :(冒号)变成%3A ； (分号)变成%3B =(等于)变成%3D ？ (问号)变成%3F @(商业A / At)变成%40

换句话说，谈到URL编码，您应该遵循RFC 1738协议。

Swift不包括+字符的编码，但它可以很好地使用这三个@:?识字课。

因此，要正确编码每个参数，.urlHostAllowed选项是不够的，你还应该添加特殊字符，例如:

encodedParameter = parameter.replacingOccurrences(of: "+", with: "%2B")

希望这能帮助那些变得疯狂的人搜索这些信息。