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。

对于Swift 5到endcode字符串

func escape(string: String) -> String {
    let allowedCharacters = string.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: ":=\"#%/<>?@\\^`{|}").inverted) ?? ""
    return allowedCharacters
}

如何使用?

let strEncoded = self.escape(string: "http://www.edamam.com/ontologies/edamam.owl#recipe_e2a1b9bf2d996cbd9875b80612ed9aa4")
print("escapedString: \(strEncoded)")

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/。

Swift 4和5(谢谢@sumizome的建议。感谢@FD_和@derickito的测试)

var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)

斯威夫特3

let allowedQueryParamAndKey =  NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)

Swift 2.2(借用Zaph's，修正url查询键和参数值)

var allowedQueryParamAndKey =  NSCharacterSet(charactersInString: ";/?:@&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)

例子:

let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"

以下是陈布莱恩回答的一个简短版本。我猜urlQueryAllowed允许控制字符通过，这很好，除非它们在查询字符串中形成键或值的一部分，在这一点上它们需要转义。

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

我自己也需要这个，所以我写了一个字符串扩展，既允许URLEncoding字符串，以及更常见的最终目标，将参数字典转换为“GET”风格的URL参数:

extension String {
    func URLEncodedString() -> String? {
        var escapedString = self.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
        return escapedString
    }
    static func queryStringFromParameters(parameters: Dictionary<String,String>) -> String? {
        if (parameters.count == 0)
        {
            return nil
        }
        var queryString : String? = nil
        for (key, value) in parameters {
            if let encodedKey = key.URLEncodedString() {
                if let encodedValue = value.URLEncodedString() {
                    if queryString == nil
                    {
                        queryString = "?"
                    }
                    else
                    {
                        queryString! += "&"
                    }
                    queryString! += encodedKey + "=" + encodedValue
                }
            }
        }
        return queryString
    }
}

享受吧!