ios 带参数的 Swift GET 请求
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Swift GET request with parameters
提问by MrSSS16
I'm very new to swift, so I will probably have a lot of faults in my code but what I'm trying to achieve is send a GETrequest to a localhost server with paramters. More so I'm trying to achieve it given my function take two parameters baseURL:string,params:NSDictionary. I am not sure how to combine those two into the actual URLRequest ? Here is what I have tried so far
我对 swift 很陌生，所以我的代码中可能会有很多错误，但我想要实现的是向GET带有参数的本地主机服务器发送请求。更重要的是，鉴于我的函数采用两个参数，我正试图实现它baseURL:string,params:NSDictionary。我不确定如何将这两者结合到实际的 URLRequest 中？这是我迄今为止尝试过的
func sendRequest(url:String,params:NSDictionary){
let urls: NSURL! = NSURL(string:url)
var request = NSMutableURLRequest(URL:urls)
request.HTTPMethod = "GET"
var data:NSData! = NSKeyedArchiver.archivedDataWithRootObject(params)
request.HTTPBody = data
println(request)
var session = NSURLSession.sharedSession()
var task = session.dataTaskWithRequest(request, completionHandler:loadedData)
task.resume()
}
}
func loadedData(data:NSData!,response:NSURLResponse!,err:NSError!){
if(err != nil){
println(err?.description)
}else{
var jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: nil) as NSDictionary
println(jsonResult)
}
}
回答by Rob
When building a GETrequest, there is no body to the request, but rather everything goes on the URL. To build a URL (and properly percent escaping it), you can also use URLComponents.
构建GET请求时，请求没有正文，而是所有内容都在 URL 上。要构建 URL（并正确地对其进行百分比转义），您还可以使用URLComponents.
var url = URLComponents(string: "https://www.google.com/search/")!
url.queryItems = [
URLQueryItem(name: "q", value: "War & Peace")
]
The only trick is that most web services need +character percent escaped (because they'll interpret that as a space character as dictated by the application/x-www-form-urlencodedspecification). But URLComponentswill not percent escape it. Apple contends that +is a valid character in a query and therefore shouldn't be escaped. Technically, they are correct, that it is allowed in a query of a URI, but it has a special meaning in application/x-www-form-urlencodedrequests and really should not be passed unescaped.
唯一的技巧是大多数 Web 服务都需要+转义字符百分比（因为它们会将其解释为application/x-www-form-urlencoded规范所规定的空格字符）。但URLComponents不会百分百逃脱它。Apple 认为这+是查询中的有效字符，因此不应转义。从技术上讲，它们是正确的，它在 URI 的查询中是允许的，但它在application/x-www-form-urlencoded请求中具有特殊含义，并且真的不应该不转义地传递。
Apple acknowledges that we have to percent escaping the +characters, but advises that we do it manually:
Apple 承认我们必须对+字符进行百分比转义，但建议我们手动进行：
var url = URLComponents(string: "https://www.wolframalpha.com/input/")!
url.queryItems = [
URLQueryItem(name: "i", value: "1+2")
]
url.percentEncodedQuery = url.percentEncodedQuery?.replacingOccurrences(of: "+", with: "%2B")
This is an inelegant work-around, but it works, and is what Apple advises if your queries may include a +character and you have a server that interprets them as spaces.
这是一个不雅的解决方法，但它有效，如果您的查询可能包含一个+字符并且您有一个将它们解释为空格的服务器，Apple 会建议您这样做。
So, combining that with your sendRequestroutine, you end up with something like:
因此，将其与您的sendRequest日常工作相结合，您最终会得到以下结果：
func sendRequest(_ url: String, parameters: [String: String], completion: @escaping ([String: Any]?, Error?) -> Void) {
var components = URLComponents(string: url)!
components.queryItems = parameters.map { (key, value) in
URLQueryItem(name: key, value: value)
}
components.percentEncodedQuery = components.percentEncodedQuery?.replacingOccurrences(of: "+", with: "%2B")
let request = URLRequest(url: components.url!)
let task = URLSession.shared.dataTask(with: request) { data, response, error in
guard let data = data, // is there data
let response = response as? HTTPURLResponse, // is there HTTP response
(200 ..< 300) ~= response.statusCode, // is statusCode 2XX
error == nil else { // was there no error, otherwise ...
completion(nil, error)
return
}
let responseObject = (try? JSONSerialization.jsonObject(with: data)) as? [String: Any]
completion(responseObject, nil)
}
task.resume()
}
And you'd call it like:
你会这样称呼它：
sendRequest("someurl", parameters: ["foo": "bar"]) { responseObject, error in
guard let responseObject = responseObject, error == nil else {
print(error ?? "Unknown error")
return
}
// use `responseObject` here
}
Personally, I'd use JSONDecodernowadays and return a custom structrather than a dictionary, but that's not really relevant here. Hopefully this illustrates the basic idea of how to percent encode the parameters into the URL of a GET request.
就个人而言，我JSONDecoder现在会使用并返回自定义struct而不是字典，但这在这里并不重要。希望这说明了如何将参数百分比编码到 GET 请求的 URL 中的基本思想。
See previous revision of this answerfor Swift 2 and manual percent escaping renditions.
请参阅此答案的先前修订版以了解 Swift 2 和手动转义百分比。
回答by Ben-Hur Batista
Use NSURLComponents to build your NSURL like this
使用 NSURLComponents 像这样构建你的 NSURL
var urlComponents = NSURLComponents(string: "https://www.google.de/maps/")!
urlComponents.queryItems = [
NSURLQueryItem(name: "q", value: String(51.500833)+","+String(-0.141944)),
NSURLQueryItem(name: "z", value: String(6))
]
urlComponents.URL // returns https://www.google.de/maps/?q=51.500833,-0.141944&z=6
font: https://www.ralfebert.de/snippets/ios/encoding-nsurl-get-parameters/
字体：https: //www.ralfebert.de/snippets/ios/encoding-nsurl-get-parameters/
回答by anoop4real
I am using this, try it in playground. Define the base urls as Struct in Constants
我正在使用这个，在操场上试试。将基本网址定义为常量中的结构
struct Constants {
struct APIDetails {
static let APIScheme = "https"
static let APIHost = "restcountries.eu"
static let APIPath = "/rest/v1/alpha/"
}
}
private func createURLFromParameters(parameters: [String:Any], pathparam: String?) -> URL {
var components = URLComponents()
components.scheme = Constants.APIDetails.APIScheme
components.host = Constants.APIDetails.APIHost
components.path = Constants.APIDetails.APIPath
if let paramPath = pathparam {
components.path = Constants.APIDetails.APIPath + "\(paramPath)"
}
if !parameters.isEmpty {
components.queryItems = [URLQueryItem]()
for (key, value) in parameters {
let queryItem = URLQueryItem(name: key, value: "\(value)")
components.queryItems!.append(queryItem)
}
}
return components.url!
}
let url = createURLFromParameters(parameters: ["fullText" : "true"], pathparam: "IN")
//Result url= https://restcountries.eu/rest/v1/alpha/IN?fullText=true
回答by Danut Pralea
Swift 3:
斯威夫特 3：
extension URL {
func getQueryItemValueForKey(key: String) -> String? {
guard let components = NSURLComponents(url: self, resolvingAgainstBaseURL: false) else {
return nil
}
guard let queryItems = components.queryItems else { return nil }
return queryItems.filter {
var originalFilename = ""
if let url = info[UIImagePickerControllerReferenceURL] as? URL, let imageIdentifier = url.getQueryItemValueForKey(key: "id") {
originalFilename = imageIdentifier + ".png"
print("file name : \(originalFilename)")
}
.name.lowercased() == key.lowercased()
}.first?.value
}
}
I used it to get the image name for UIImagePickerControllerin func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]):
我用它来获取UIImagePickerControllerin的图像名称func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any])：
extension Dictionary where Key : CustomStringConvertible, Value : CustomStringConvertible {
func stringFromHttpParameters() -> String {
var parametersString = ""
for (key, value) in self {
parametersString += key.description + "=" + value.description + "&"
}
return parametersString
}
}
回答by Reza Shirazian
You can extend your Dictionaryto only provide stringFromHttpParameterif both key and value conform to CustomStringConvertablelike this
您可以将您的扩展扩展Dictionary为仅stringFromHttpParameter在键和值都符合CustomStringConvertable这样的情况下提供
extension Dictionary {
/// Build string representation of HTTP parameter dictionary of keys and objects
///
/// :returns: String representation in the form of key1=value1&key2=value2 where the keys and values are percent escaped
func stringFromHttpParameters() -> String {
var parametersString = ""
for (key, value) in self {
if let key = key as? String,
let value = value as? String {
parametersString = parametersString + key + "=" + value + "&"
}
}
parametersString = parametersString.substring(to: parametersString.index(before: parametersString.endIndex))
return parametersString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!
}
}
this is much cleaner and prevents accidental calls to stringFromHttpParameterson dictionaries that have no business calling that method
这更干净，并防止意外调用stringFromHttpParameters没有业务调用该方法的字典
回答by etayluz
This extension that @Rob suggested works for Swift 3.0.1
@Rob 建议的这个扩展适用于 Swift 3.0.1
I wasn't able to compile the version he included in his post with Xcode 8.1 (8B62)
我无法使用 Xcode 8.1 (8B62) 编译他在帖子中包含的版本
let dictionary = ["method":"login_user",
"cel":mobile.text!
"password":password.text!] as Dictionary<String,String>
for (key, value) in dictionary {
data=data+"&"+key+"="+value
}
request.HTTPBody = data.dataUsingEncoding(NSUTF8StringEncoding);
回答by Fabio Guerra
I use:
我用：
##代码##