Swift 3和Swift 4中的WebService API方法？ [英] WebService API Method in Swift 3 and Swift 4?

问题描述

我是iOS Swift 的新手，我想在单独的类中创建单独的方法（例如 NSObject ） Web服务，以便我可以在任何ViewController中使用它，并使用 NSURLSession 和<$解析任何类型的 JSON 响应c $ c> Alamofire 。

I am new to Swift iOS and i want to create a separate method in separate class(like NSObject) of web services so that i can use it in any ViewController and parse any type of JSON response using NSURLSession and Alamofire . Could someone help me.

推荐答案

class WebRequester: NSObject {

    static let shared = WebRequester()

    let session = URLSession.shared

    func request(urlStr:String, parameter:String, token:String? = nil, callback:@escaping (_ result:NSDictionary?, error:Error?) -> Void) {

        let url = URL(string: BaseURL + urlStr)

        debugPrint("=====================")
        debugPrint(url ?? "")
        debugPrint(parameter)
        debugPrint("=====================")

        var request = URLRequest(url: url!)
        request.httpMethod = "POST"
        request.httpBody = parameter.data(using: String.Encoding.utf8)


        print("Token :", (token ?? ""))
        request.setValue(token, forHTTPHeaderField: "Authorization")


        let task = session.dataTask(with: request) { (data, response, error) in
            DispatchQueue.main.async {
                if error == nil {
                   do {
                         let jsonObj = try JSONSerialization.jsonObject(with: data!, options: JSONSerialization.ReadingOptions.allowFragments)
                         if let dic = jsonObj as? NSDictionary {
                             callback(dic, error)
                         }
                   }
                   catch {
                      callback(nil, error)
                   }
                }
                else {
                   callback(nil, error)
                }

            }
        }
        task.resume()
    }

}

您需要将参数作为字符串传递

You need to pass parameter as string

var params = "user_id=" + "12"
params += "&course_id=" + "1"

WebRequester.shared.request(urlStr: urlStr, parameter: params) { (result, error) in
    DispatchQueue.main.async {
         print(result)
    }
}

您也可以将参数作为字典传递，但需要使用以下字典扩展

you can also pass parameter as dictionary but need to convert in string using following Dictionary extension

request.httpBody = parameter.stringFromHttpParameters（）。data（使用：
String.Encoding.utf8）

request.httpBody = parameter.stringFromHttpParameters().data(using: String.Encoding.utf8)

extension Dictionary {
    func stringFromHttpParameters() -> String {
        let parameterArray = self.map { (key, value) -> String in
            let percentEscapedKey = (key as! String).addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!
            let percentEscapedValue = (value as AnyObject).addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!
            return "\(percentEscapedKey)=\(percentEscapedValue)"
        }

        return parameterArray.joined(separator: "&")
    }
}

WebRequester

• Http请求


• 具有单个图像的多部分


• 具有多个图像的多部分

您需要根据api响应更改响应结构

You need to change response structure as per your api response

您需要设置授权和api根据您的api键

you need to set authorisation and api key as per your api

WebRequester

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