Swift - 编码URL-Java 学习之路

220

如果我编码这样的字符串：

var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

它没有逃脱斜线 / .

我搜索并找到了这个Objective C代码：

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

是否有更简单的方法来编码URL，如果没有，我如何在Swift中编写它？

11 回答

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)

result:

"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)

result:

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

回复于 2024-04-26T14:34:25+08:00

0
斯威夫特4：

它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)!
```
在这个有用的tool之后，你应该保证为你的参数编码这些字符：
- $（美元符号）变为％24
- ＆（＆符号）变为％26
- （加号）成为％2B
- ，（逗号）变为％2C
- :(冒号）变为％3A
- ; （半结肠）变为％3B
- =（等于）变为％3D
- ？（问号）变为％3F
- @（商业A / At）成为％40
换句话说，关于URL编码，您应该遵循RFC 1738 protocol .

And Swift don't cover the encoding of the + char for example ，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 .
回复于 2024-04-26T14:34:25+08:00

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
    }
}

请享用！

回复于 2024-04-26T14:34:25+08:00

482
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!)"
```
使用URL编码的任何标准字符集（如 URLQueryAllowedCharacterSet 或 URLHostAllowedCharacterSet ）将不起作用，因为它们不排除 = 或 & 字符 .

Note 通过使用 .alphanumerics ，它将编码一些不需要编码的字符（如 - ， . ， _ 或 ~ - 参见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!)"
```
Warning: encoded 参数强制解包 . 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? . 而不是强行展开 encoded! ，您可以使用 encoded ?? "" 或使用 if let encoded = ... .
回复于 2024-04-26T14:34:25+08:00

斯威夫特3

在Swift 3中有 addingPercentEncoding

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

输出：

测试％2Ftest

斯威夫特1

在iOS 7及更高版本中有 stringByAddingPercentEncodingWithAllowedCharacters

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

输出：

测试％2Ftest

以下是有用（反向）字符集：

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

如果要转义不同的字符集，请创建一个集合：
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％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)\'")
}

回复于 2024-04-26T14:34:25+08:00

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
}

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

回复于 2024-04-26T14:34:25+08:00

一切都是一样的

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

// test%2Ftest

回复于 2024-04-26T14:34:25+08:00

斯威夫特3：

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

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

回复于 2024-04-26T14:34:25+08:00

Swift 4 (not tested - please comment if it works or not. Thanks @sumizome for suggestion)

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 （从Zaph借用并更正url查询键和参数值）

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

Example:

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"

这是一个较短版本的Bryan Chen 's answer. I' d猜测 urlQueryAllowed 允许控制字符通过哪个很好，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 .

回复于 2024-04-26T14:34:25+08:00

9
这在Swift 4.2中对我有用 . 用例是从剪贴板或类似的URL获取一个URL，该URL可能已经有转义字符，但也包含可能导致 URL(string:) 失败的Unicode字符 .
```
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)
```
最后 url 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，保留 %20 和 + 间距，对Unicode字符进行编码，并且原始 urlText 中的 %20 不进行双重编码 .

Caveat: 这是一个快捷方式，因为它在整个URL上使用 .urlQueryAllowed ;它还假设URL已在浏览器中运行 . 明智地使用！我也愿意接受建议（我没有用 URLComponents 斩断整个网址，应用单个字符集，并重新组装） .
回复于 2024-04-26T14:34:25+08:00

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, host: String, path: String, queryItems: [URLQueryItem]) {
        self.init()
        self.scheme = scheme
        self.host = host
        self.path = path
        self.queryItems = queryItems
    }
}

if let url = URLComponents(scheme: "https",
                             host: "www.google.com",
                             path: "/search",
                       queryItems: [URLQueryItem(name: "q", value: "Formula One")]).url {

    print(url)  // https://www.google.com/search?q=Formula%20One
}

回复于 2024-04-26T14:34:25+08:00

Swift - 编码URL

11 回答

斯威夫特3：

斯威夫特4：

斯威夫特4

斯威夫特3

斯威夫特1

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