Swift - 编码 URL

斯威夫特 3

在 Swift 3 中有 addingPercentEncoding

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

输出：

测试%2F测试

斯威夫特 1

在 iOS 7 及更高版本中有 stringByAddingPercentEncodingWithAllowedCharacters

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

输出：

测试%2F测试

以下是有用的（反转的）字符集：

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

如果您想转义一组不同的字符，请创建一组：添加“=”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%2F测试%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 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
}

---

斯威夫特 4 & 5

要对 URL 中的参数进行编码，我发现使用 .alphanumerics 字符集是最简单的选项：

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

使用任何标准的 URL 编码字符集（如 URLQueryAllowedCharacterSet 或 URLHostAllowedCharacterSet）都不起作用，因为它们不排除 = 或 & 字符。

注意，使用 .alphanumerics 会编码一些不需要编码的字符（如 -、.、_ 或 ~ - 参见 2.3 . RFC 3986 中的未保留字符）。我发现使用 .alphanumerics 比构建自定义字符集更简单，并且不介意要编码一些额外的字符。如果这让您感到困扰，请按照 How to percent encode a URL String 中的说明构建自定义字符集，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 参数被强制解包。对于无效的 unicode 字符串，它可能会崩溃。请参阅Why is the return value of String.addingPercentEncoding() optional?。您可以使用 urlEncoded ?? "" 或 if let urlEncoded = urlEncoded { ... }，而不是强制展开 urlEncoded!。

---

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

---

斯威夫特 3：

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

1.编码查询：

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.编码网址：

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"

---

Swift 4 & 5（感谢@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 并更正 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"

这是 Bryan Chen 答案的简短版本。我猜想 urlQueryAllowed 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，此时它们需要转义。

---

斯威夫特 4：

这取决于您的服务器遵循的编码规则。

Apple 提供了此类方法，但它没有报告它遵循的那种 RCF 协议。

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

按照这个有用的 tool，您应该保证这些字符的编码用于您的参数：

$（美元符号）变为 %24

&（与号）变为 %26

+（加号）变为 %2B

,（逗号）变为 %2C

:（冒号）变为 %3A

; （分号）变为 %3B

=（等于）变为 %3D

? （问号）变为 %3F

@（商业 A / At）变为 %40

换句话说，说到 URL 编码，您应该遵循 RFC 1738 protocol。

例如，Swift 没有涵盖 + char 的编码，但它适用于这三个 @ : ？字符。

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

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

希望这有助于疯狂搜索这些信息的人。

---

一切都一样

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

// test%2Ftest

---

斯威夫特 4.2

一种快速的单线解决方案。将 originalString 替换为您要编码的字符串。

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

Online Playground Demo

---

这在 Swift 5 中对我有用。用例是从剪贴板或类似的 URL 获取可能已经有转义字符但也包含可能导致 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，使用对 URL 进行编码的方法扩展 String：

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)

末尾的 url 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

请注意，保留 %20 和 + 间距，对 Unicode 字符进行编码，原始 urlText 中的 %20 不是双重编码的，并且锚点（片段或 #）保留。

编辑：现在检查每个组件的有效性。

---

对于 Swift 5 来结束编码字符串

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)")

---

我自己也需要这个，所以我写了一个字符串扩展，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
    }
}

享受！

---

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
}

我从这个链接找到了上面的函数：http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/。

---

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

---

帮助我的是我创建了一个单独的 NSCharacterSet 并将其用于 UTF-8 编码的字符串，即 textToEncode 以生成所需的结果：

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)"

---

这些答案都不适合我。当 url 包含非英文字符时，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)

根据您尝试执行的参数，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 -._~/?%$!:

---

斯威夫特 4.2

有时发生这种情况只是因为 slug 中有空间或缺少通过 API URL 传递的参数的 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。

---

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  

---

Swift 5 如果你想像下面这样编码字符串，你可以尝试 .afURLQueryAllowed 选项

let testString = "6hAD9/RjY+SnGm&B"
let escodedString = testString.addingPercentEncoding(withAllowedCharacters: .afURLQueryAllowed)
print(escodedString!) 

//编码后的字符串就像 en6hAD9%2FRjY%2BSnGm%26B

---