iOS 中的 URL 解码
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URL decode in iOS
提问by JibW
I am using Swift 1.2 to develop my iPhone application and I am communicating with a http web service.
我正在使用 Swift 1.2 来开发我的 iPhone 应用程序，并且我正在与一个 http Web 服务进行通信。
The response I am getting is in query string format (key-value pairs) and URL encoded in .Net.
我得到的响应是查询字符串格式（键值对）和以.Net.
I can get the response, but looking the proper way to decode using Swift.
我可以得到响应，但正在寻找使用 Swift 进行解码的正确方法。
Sample response is as follows
示例响应如下
status=1&message=The+transaction+for+GBP+12.50+was+successful
Tried following way to decode and get the server response
尝试以下方式解码并获取服务器响应
// This provides encoded response String
var responseString = NSString(data: data, encoding: NSUTF8StringEncoding) as! String
var decodedResponse = responseString.stringByReplacingEscapesUsingEncoding(NSUTF8StringEncoding)!
How can I replace all URL escaped characters in the string?
如何替换字符串中的所有 URL 转义字符？
回答by Vyacheslav
To encode and decode urls create this extention somewhere in the project:
要对 url 进行编码和解码，请在项目中的某处创建此扩展名：
Swift 2.0
斯威夫特 2.0
extension String
{
func encodeUrl() -> String
{
return self.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())
}
func decodeUrl() -> String
{
return self.stringByRemovingPercentEncoding
}
}
Swift 3.0
斯威夫特 3.0
extension String
{
func encodeUrl() -> String
{
return self.addingPercentEncoding( withAllowedCharacters: NSCharacterSet.urlQueryAllowed())
}
func decodeUrl() -> String
{
return self.removingPercentEncoding
}
}
Swift 4.1
斯威夫特 4.1
extension String
{
func encodeUrl() -> String?
{
return self.addingPercentEncoding( withAllowedCharacters: NSCharacterSet.urlQueryAllowed)
}
func decodeUrl() -> String?
{
return self.removingPercentEncoding
}
}
回答by u5150587
Swift 2 and later (Xcode 7)
Swift 2 及更高版本 (Xcode 7)
var s = "aa bb -[:/?&=;+!@#$()',*]";
let sEncode = s.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())
let sDecode = sEncode?.stringByRemovingPercentEncoding
回答by matt
The stringByReplacingEscapesUsingEncodingmethod is behaving correctly. The "+"character is not part of percent-encoding. This server is using it incorrectly; it should be using a percent-escaped space here (%20). If, for a particular response, you want spaces where you see "+"characters, you just have to work around the server behavior by performing the substitution yourself, as you are already doing.
该stringByReplacingEscapesUsingEncoding方法行为正确。该"+"字符不是百分比编码的一部分。该服务器使用不当；它应该在此处使用百分比转义空格 ( %20)。如果对于特定的响应，您想要在其中看到"+"字符的空间，您只需像您已经在做的那样，通过自己执行替换来解决服务器行为。
回答by Cai?ara
In my case, I NEED a plus ("+") signal in a phone number in parameters of a query string, like "+55 11 99999-5555". After I discovered that the swift3 (xcode 8.2) encoder don't encode "+" as plus signal, but space, I had to appeal to a workaround after the encode:
就我而言，我需要在查询字符串参数中的电话号码中添加加号（“+”）信号，例如“+55 11 99999-5555”。在我发现 swift3 (xcode 8.2) 编码器不会将“+”编码为加号信号，而是将空格编码为空格后，我不得不在编码后寻求解决方法：
Swift 3.0
斯威夫特 3.0
_strURL = _strURL.replacingOccurrences(of: "+", with: "%2B")
回答by Amul4608
In Swift 3
在斯威夫特 3
extension URL {
var parseQueryString: [String: String] {
var results = [String: String]()
if let pairs = self.query?.components(separatedBy: "&"), pairs.count > 0 {
for pair: String in pairs {
if let keyValue = pair.components(separatedBy: "=") as [String]? {
results.updateValue(keyValue[1], forKey: keyValue[0])
}
}
}
return results
}
}
in your code to access below
在您的代码中访问下面
let parse = url.parseQueryString
print("parse \(parse)" )
回答by nikans
It's better to use built-in URLComponentsstruct, since it follows proper guidelines.
最好使用内置URLComponents结构，因为它遵循适当的准则。
extension URL
{
var parameters: [String: String?]?
{
if let components = URLComponents(url: self, resolvingAgainstBaseURL: false),
let queryItems = components.queryItems
{
var parameters = [String: String?]()
for item in queryItems {
parameters[item.name] = item.value
}
return parameters
} else {
return nil
}
}
}
回答by ScottyBlades
You only need:
你只需要：
print("Decode: ", yourUrlAsString.removingPercentEncoding)
