带参数的 Swift GET 请求

Question

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带有参数的本地主机服务器发送 GET 请求。更重要的是，鉴于我的函数采用两个参数baseURL:string,params:NSDictionary，我正在尝试实现它。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)

    }

}

Answer 1

在构建GET 请求时，请求没有正文，而是所有内容都在 URL 上。要构建 URL（并适当地转义它），您还可以使用 URLComponents。

var url = URLComponents(string: "https://www.google.com/search/")!

url.queryItems = [
    URLQueryItem(name: "q", value: "War & Peace")
]

唯一的技巧是大多数网络服务需要+ 字符百分比转义（因为他们会将其解释为application/x-www-form-urlencoded specification 规定的空格字符）。但是URLComponents 不会逃脱它。 Apple 认为 + 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，实际上不应该不转义地传递。

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")

这是一种不优雅的解决方法，但它有效，如果您的查询可能包含 + 字符并且您有一个将它们解释为空格的服务器，Apple 建议这样做。

因此，将其与您的 sendRequest 例程结合起来，您最终会得到如下结果：

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                                  // was there no error
        else {
            completion(nil, error)
            return
        }
        
        let responseObject = (try? JSONSerialization.jsonObject(with: data)) as? [String: Any]
        completion(responseObject, nil)
    }
    task.resume()
}

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
}

就个人而言，我现在会使用 JSONDecoder 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。

---

请参阅previous revision of this answer 了解 Swift 2 和手动百分比转义再现。

Answer 2

使用 NSURLComponents 构建你的 NSURL 像这样

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

字体：https://www.ralfebert.de/snippets/ios/encoding-nsurl-get-parameters/

Answer 3

我正在使用这个，在操场上试试。将基本 url 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

Answer 4

Swift 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 {
                 $0.name.lowercased() == key.lowercased()
                 }.first?.value
    }
}

我用它在func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) 中获取UIImagePickerController 的图像名称：

var originalFilename = ""
if let url = info[UIImagePickerControllerReferenceURL] as? URL, let imageIdentifier = url.getQueryItemValueForKey(key: "id") {
    originalFilename = imageIdentifier + ".png"
    print("file name : \(originalFilename)")
}

Answer 5

如果键和值都符合 CustomStringConvertable 像这样，您可以扩展您的 Dictionary 以仅提供 stringFromHttpParameter

extension Dictionary where Key : CustomStringConvertible, Value : CustomStringConvertible {
  func stringFromHttpParameters() -> String {
    var parametersString = ""
    for (key, value) in self {
      parametersString += key.description + "=" + value.description + "&"
    }
    return parametersString
  }
}

这样更简洁，可以防止在没有业务调用该方法的字典上意外调用 stringFromHttpParameters

Answer 6

@Rob 建议的这个扩展适用于 Swift 3.0.1

我无法使用 Xcode 8.1 (8B62) 编译他在帖子中包含的版本

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)!
    }

}

Answer 7

我用：

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);