Swift-编码URL

如果我编码这样的字符串：

var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

它不能逃脱斜线 /。

我搜索并找到了此Objective C代码：

NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
                        NULL,
                        (CFStringRef)unencodedString,
                        NULL,
                        (CFStringRef)@"!*'();:@&=+$,/?%#[]",
                        kCFStringEncodingUTF8 );

有没有更简单的方法来编码URL，否则，如何在Swift中编写此代码？

迅捷3

在Swift 3中有 addingPercentEncoding

let originalString = "test/test"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(escapedString!)

输出：

测试％2Ftest

斯威夫特1

在iOS 7及更高版本中， stringByAddingPercentEncodingWithAllowedCharacters

var originalString = "test/test"
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
println("escapedString: \(escapedString)")

输出：

测试％2Ftest

以下是有用的（倒置）字符集：

URLFragmentAllowedCharacterSet  "#%<>[\]^`{|}
URLHostAllowedCharacterSet      "#%/<>?@\^`{|}
URLPasswordAllowedCharacterSet  "#%/:<>?@[\]^`{|}
URLPathAllowedCharacterSet      "#%;<>?[\]^`{|}
URLQueryAllowedCharacterSet     "#%<>[\]^`{|}
URLUserAllowedCharacterSet      "#%/:<>?@[\]^`

如果要转义其他字符集，请创建一个字符集：
添加了“ =”字符的示例：

var originalString = "test/test=42"
var customAllowedSet =  NSCharacterSet(charactersInString:"=\"#%/<>?@\\^`{|}").invertedSet
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(customAllowedSet)
println("escapedString: \(escapedString)")

输出：

测试％2Ftest％3D42

验证不在集合中的ASCII字符的示例：

func printCharactersInSet(set: NSCharacterSet) {
    var characters = ""
    let iSet = set.invertedSet
    for i: UInt32 in 32..<127 {
        let c = Character(UnicodeScalar(i))
        if iSet.longCharacterIsMember(i) {
            characters = characters + String(c)
        }
    }
    print("characters not in set: \'\(characters)\'")
}

您可以使用URLComponents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
}

斯威夫特3：

let originalString = "http://www.ihtc.cc?name=htc&title=iOS开发工程师"

1. encodingQuery：

let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:NSCharacterSet.urlQueryAllowed)

结果：

"http://www.ihtc.cc?name=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88" 

2. encodingURL：

let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)

结果：

"http:%2F%2Fwww.ihtc.cc%3Fname=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"

斯威夫特3：

let allowedCharacterSet = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted)

if let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet) {
//do something with escaped string
}

斯威夫特4

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

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

无法使用任何标准字符集进行URL编码（如URLQueryAllowedCharacterSet或URLHostAllowedCharacterSet），因为它们不排除=或&字符。

注意，通过使用.alphanumerics它编码无需编码的某些字符（如-，.，_或~-见2.3未保留的字符。在RFC 3986）。我发现使用的方法.alphanumerics比构造自定义字符集更简单，并且不介意要编码的其他一些字符。如果那让您感到困扰，请按照如何对URL字符串进行百分比编码中所述构造一个自定义字符集，例如：

var allowed = CharacterSet.alphanumerics
allowed.insert(charactersIn: "-._~") // as per RFC 3986
let encoded = parameter.addingPercentEncoding(withAllowedCharacters: allowed)
let url = "http://www.example.com/?name=\(encoded!)"

警告：该encoded参数已强制展开。对于无效的unicode字符串，它可能会崩溃。请参见为什么String.addingPercentEncoding（）的返回值是可选的？。不用强制展开，encoded!您可以使用encoded ?? ""或使用if let encoded = ...。

一切都一样

var str = CFURLCreateStringByAddingPercentEscapes(
    nil,
    "test/test",
    nil,
    "!*'();:@&=+$,/?%#[]",
    CFStringBuiltInEncodings.UTF8.rawValue
)

// test%2Ftest

斯威夫特4：

这取决于服务器遵循的编码规则。

Apple提供了此类方法，但没有报告遵循的那种RCF协议。

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

使用此有用的工具后，您应该保证为参数使用这些字符的编码：

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

换句话说，谈到URL编码，您应该遵循RFC 1738协议。

而且Swift不会覆盖+ char的编码，例如，它与这三个@：一起使用效果很好。字符

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

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

希望这可以帮助某个变得疯狂的人搜索这些信息。

Swift 4（未经测试-请评论它是否有效。感谢@sumizome的建议）

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

迅捷3

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

Swift 2.2（从Zaph借用并更正url查询键和参数值）

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

例：

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"

这是Bryan Chen答案的简短版本。我猜想这urlQueryAllowed将允许控制字符通过，除非它们构成查询字符串中键或值的一部分，否则必须将其转义。

斯威夫特4.2

一种快速的单线解决方案。替换originalString为要编码的字符串。

var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[]{} ").inverted)

在线游乐场演示

自己需要这个，所以我写了一个String扩展名，它既允许URLEncoding字符串，也允许更常见的最终目标，将参数字典转换为“ GET”样式的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
    }
}

请享用！

这个正在为我工​​作。

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
}

我从此链接找到了上述功能：http : //useyourloaf.com/blog/how-to-percent-encode-a-url-string/。

let Url = URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) ?? "")

SWIFT 4.2

有时，发生这种情况的原因仅在于：在子标签中存在空间，或者缺少通过API URL传递参数的URL编码。

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bitmapRepresentation。

这在Swift 5中为我工作。该用例是从剪贴板或类似的网址中获取一个URL，该URL可能已经转义了字符，但还包含可能导致URLComponents或URL(string:)失败的Unicode字符。

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，扩展String一种用于编码URL的方法：

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
    }

}

可以像这样测试：

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)

最后的值url：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

请注意，保留了%20和+间距，都对Unicode字符进行了编码，%20原始字符中的urlText不会进行双重编码，并且锚点（片段或#）仍然保留。

编辑：现在检查每个组件的有效性。

对于Swift 5结束代码字符串

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)")

这些答案都不对我有用。网址包含非英语字符时，我们的应用程序崩溃了。

 let unreserved = "-._~/?%$!:"
 let allowed = NSMutableCharacterSet.alphanumeric()
     allowed.addCharacters(in: unreserved)

 let escapedString = urlString.addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)

根据您要执行的操作的参数，您可能只想创建自己的字符集。以上允许使用英文字符，并且-._~/?%$!: