Swift - 编码URL

时间：2014-07-03 10:56:03

标签： ios swift urlencode

如果我对这样的字符串进行编码：

var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

它不会逃避斜杠/。

我已经搜索过并找到了这个Objective C代码：

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

是否有更简单的方法对URL进行编码，如果没有，我该如何在Swift中编写？

19 个答案:

答案 0 :(得分：527)

Swift 3

在Swift 3中有addingPercentEncoding 

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

输出：

  

测试％2Ftest

Swift 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％2Ftest

以下是有用的（反向）字符集：

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

如果您想要转义不同的字符集，请创建一个集：
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)\'")
}

答案 1 :(得分：46)

您可以使用URLComponents来避免手动百分比转义查询字符串：

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
}

答案 2 :(得分：32)

斯威夫特3：

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

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

答案 3 :(得分：24)

Swift 3：

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

<强> 1。 encodingQuery： 

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

<强>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"

<强> 2。 encodingURL： 

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

<强>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"

答案 4 :(得分：18)

Swift 4

要在URL中对参数进行编码，我使用.alphanumerics字符集找到最简单的选项：

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

使用任何标准字符集进行网址编码（例如URLQueryAllowedCharacterSet或URLHostAllowedCharacterSet）都不起作用，因为它们不会排除=或&个字符。

注意，使用.alphanumerics，它会对一些不需要编码的字符进行编码（例如-，.，_或~ - 请参阅 2.3。RFC 3986中未保留的字符）。我发现使用.alphanumerics比构造自定义字符集更简单，并且不介意要编码一些其他字符。如果这让您感到困扰，请构建一个自定义字符集，如How to percent encode a URL String中所述，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： encoded参数已强制解包。对于无效的unicode字符串，它可能会崩溃。见Why is the return value of String.addingPercentEncoding() optional?。您可以使用encoded!或使用encoded ?? ""来代替强行展开if let encoded = ...。

答案 5 :(得分：12)

一切都是一样的

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

// test%2Ftest

答案 6 :(得分：10)

Swift 4（未经测试 - 请评论是否有效。感谢@sumizome提出建议） 

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

Swift 3 

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

Swift 2.2 （借用Zaph＆＃39;并更正网址查询键和参数值）

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

示例： 

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"

这是布莱恩·陈的答案的缩短版。我猜测urlQueryAllowed允许控制字符是正常的，除非它们构成查询字符串中的键或值的一部分，此时它们需要被转义。

答案 7 :(得分：8)

斯威夫特4：

这取决于您的服务器遵循的编码规则。

Apple提供此类方法，但它没有报告它遵循的RCF协议。

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

遵循这个有用的tool，你应该保证为你的参数编码这些字符：

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

换句话说，谈到网址编码，您应该遵循RFC 1738 protocol。

Swift并未涵盖+ 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")

希望这可以帮助那些疯狂的人搜索这些信息。

答案 8 :(得分：4)

我自己需要这个，所以我编写了一个String扩展，它允许URLEncoding字符串，以及更常见的最终目标，将参数字典转换为“GET”样式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！

答案 9 :(得分：2)

这个适合我。

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
}

我在此链接中找到了上述功能：http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/。

答案 10 :(得分：2)

Swift 4.2

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

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

Online Playground Demo

答案 11 :(得分：0)

SWIFT 4.2

有时发生这种情况的原因是，在通过API URL传递的参数中，是否存在空格或缺少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。

答案 12 :(得分：0)

为Swift 5终止字符串 

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)")

答案 13 :(得分：0)

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-._~/?%$!:

答案 14 :(得分：0)

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 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)"

答案 15 :(得分：-1)

这在Swift 4.2中为我工作。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Unicode字符，这可能导致URL(string:)失败。

func encodedUrl(from string: String) -> URL? {
    // Remove preexisting encoding
    guard let decodedString = string.removingPercentEncoding,
        // Reencode, to revert decoding while encoding missed characters
        let percentEncodedString = decodedString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) else {
            // Coding failed
            return nil
    }
    // Create URL from encoded string, or nil if failed
    return URL(string: percentEncodedString)
}

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

url末尾的值：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

请注意，%20和+的间距均会保留，对Unicode字符进行编码，并且原始%20中的urlText不会进行双重编码。

注意事项：这是快捷方式，因为它在整个URL上使用.urlQueryAllowed；它还假定该URL已经在浏览器中工作。明智地使用！我也乐于接受建议（我停止使用URLComponents切掉整个URL，应用单个字符集并重新组装）。

答案 16 :(得分：-1)

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!)

//编码后的字符串将类似于 en6hAD9%2FRjY%2BSnGm%26B

答案 17 :(得分：-2)

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

答案 18 :(得分：-2)

版本：Swift 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
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