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？
在Swift 3中有addingPercentEncoding
let originalString = "test/test"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(escapedString!)
输出：
测试％2Ftest
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 "#%/:<>?@[\]^`
如果要转义不同的字符集，请创建一个集合： 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)\'")
}
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/。
快速的一线解决方案。将originalString替换为要编码的String。
var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[]{} ").inverted)
SWIFT 4.2
有时这只是因为slug中存在空间或者没有对通过API URL的参数进行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。
这对我在Swift 5中起作用。用例是从剪贴板或类似物中获取一个URL，该URL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，使用一种编码URL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
}
}
哪个可以测试如下：
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)
url最后的价值：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，%20和+间距都被保留，Unicode字符被编码，原始%20中的urlText不是双重编码，并且锚（片段或#）仍然存在。
编辑：现在检查每个组件的有效性。
let Url = URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) ?? "")
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
}
斯威夫特3：
let allowedCharacterSet = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted)
if let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet) {
//do something with escaped string
}
let originalString = "http://www.ihtc.cc?name=htc&title=iOS开发工程师"
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. 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"
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）都不起作用，因为它们不排除=或&字符。
请注意，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。如果这困扰你，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?。而不是力量展开encoded!你可以使用encoded ?? ""或使用if let encoded = ...。
一切都是一样的
var str = CFURLCreateStringByAddingPercentEscapes(
nil,
"test/test",
nil,
"!*'();:@&=+$,/?%#[]",
CFStringBuiltInEncodings.UTF8.rawValue
)
// test%2Ftest
它取决于您的服务器遵循的编码规则。
Apple提供此类方法，但它没有报告它遵循的RCF协议。
var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!
在这个有用的tool之后，您应该保证为您的参数编码这些字符：
换句话说，谈到URL编码，你应该遵循RFC 1738 protocol。
例如，Swift没有涵盖+ char的编码，但它适用于这三个@：？字符。
因此，要正确编码每个参数，.urlHostAllowed选项是不够的，您还应该添加特殊字符，例如：
encodedParameter = parameter.replacingOccurrences(of: "+", with: "%2B")
希望这可以帮助那些疯狂的人搜索这些信息。
Swift 4（未经测试 - 请评论是否有效。感谢@sumizome提出建议）
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
斯威夫特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的答案的缩短版。我猜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。
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！