Question

我有网络服务

http://dev.nsol.sg/projects/sneakers/Api/get_single_news

方法：POST 必需参数：

userid

newsid

响应：

success = 0 (error), 1 (success)
message = Description of result if success = 0 then message will have the detail description
data: if success = 1 then we will have complete details of common requirements like single news
newsurl: News Image URL + Concatenate Value of img from DB

这是我尝试从网址

获取数据

func single_news(userid: Int) {

        var request = URLRequest(url: URL(string: news_url)!)
        request.httpMethod = "POST"

        //Pass your parameter here
        let postString = "userid=\(userid)"
        request.httpBody = postString.data(using: .utf8)
        let task = URLSession.shared.dataTask(with: request) { data, response, error in

            guard let data = data, error == nil else {
                print("error=(error)")
                return
            }


            let json: Any?

            do
            {

                json = try JSONSerialization.jsonObject(with: data, options: [])
                print("abcnews")
                //here is your JSON
                print(json)
            }
            catch
            {
                return
            }

            guard let server_response = json as? NSDictionary else
            {
                return
            }
        }
        task.resume()
    }

如何传递第二个参数newsid。您可以从此链接下载项目https://drive.google.com/file/d/1mBlYcNaW8K3s6Y2V6BY6zdM0Znvyw3Si/view?usp=sharing

Answer 1

重写你的功能

func single_news(userid: Int , newsid : Int) {}

我认为newsid也是Int值。

然后像这样写postString ......

let postString = "userid=\(userid)&newsid=\(newsid)"

我希望这会对你有所帮助。

Answer 2

如果我理解正确，您只想创建并发送具有两个属性的JSON对象。

你可以这样做：

let body: [String: Any] = ["userid": userid, 
                           "newsid": newsid]

request.httpBody = try! JSONSerialization.data(withJSONObject: body, options: [])

如何将两个参数传递给webservice

2 个答案: