Swift – 编码URL
如果我编码像这样的string：
var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding) 它不会逃避斜杠/ 。
我已经search并find了这个Objective C代码：
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes( NULL, (CFStringRef)unencodedString, NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8 );
有一个更简单的方法来编码的url，如果不是，我怎么写在Swift？
Swift 3
在Swift 3中addingPercentEncoding
var originalString = "test/test" var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) print(escapedString!)
输出：
testing％2Ftest
斯威夫特1
在iOS 7及以上，有stringByAddingPercentEncodingWithAllowedCharacters
var originalString = "test/test" var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet()) println("escapedString: \(escapedString)")
输出：
testing％2Ftest
以下是有用的（反转）字符集：
URLFragmentAllowedCharacterSet "#%<>[\]^`{|} URLHostAllowedCharacterSet "#%/<>?@\^`{|} URLPasswordAllowedCharacterSet "#%/:<>?@[\]^`{|} URLPathAllowedCharacterSet "#%;<>?[\]^`{|} URLQueryAllowedCharacterSet "#%<>[\]^`{|} URLUserAllowedCharacterSet "#%/:<>?@[\]^`
如果你想要不同的字符集转义创build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：
testing％2Ftest％3D42
validation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)\'") }
Swift 3：
let allowedCharacterSet = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted) if let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet) { //do something with escaped string }
您可以使用URLComponents来避免手动百分号转义您的查询string：
let scheme = "https" let host = "www.google.com" let path = "/search" let queryItem = URLQueryItem(name: "q", value: "Formula One") var urlComponents = URLComponents() urlComponents.scheme = scheme urlComponents.host = host urlComponents.path = path urlComponents.queryItems = [queryItem] if let url = urlComponents.url { print(url) // "https://www.google.com/search?q=Formula%20One" }
extension URLComponents { init(scheme: String, host: String, path: String, queryItems: [URLQueryItem]) { self.init() self.scheme = scheme self.host = host self.path = path self.queryItems = queryItems } } if let url = URLComponents(scheme: "https", host: "www.google.com", path: "/search", queryItems: [URLQueryItem(name: "q", value: "Formula One")]).url { print(url) // https://www.google.com/search?q=Formula%20One }
一切都是一样的
var str = CFURLCreateStringByAddingPercentEscapes( nil, "test/test", nil, "!*'();:@&=+$,/?%#[]", CFStringBuiltInEncodings.UTF8.rawValue ) // test%2Ftest
Swift 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"
Swift 3
let allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:@&=+$, ") paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 2.2 （从Zaph中借用并更正url查询键和参数值）
var allowedQueryParamAndKey = NSCharacterSet(charactersInString: ";/?:@&=+$, ").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"
这是布赖恩·陈的答案的一个更短的版本。 我猜想urlQueryAllowed是允许控制字符，通过它是好的，除非他们形成键或值的一部分，在您的查询string，他们需要逃脱。
自己需要这个，所以我写了一个String扩展，它既允许URLEncodingstring，也允许更常见的最终目标，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 } }
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 }
我从这个链接find上面的函数： http : //useyourloaf.com/blog/how-to-percent-encode-a-url-string/ 。
