如何在Swift iOS中使模型类遵循JSON响应 [英] How to make model class for following JSON response in swift iOS

问题描述

您好，我是swift ios的初学者，我的要求是必须显示对表列表的Json响应，我从网络服务获得了响应，并且响应如下所示

Hi i am beginner for swift ios and my requirement is have to display Json response to table list i got response from web-services and response seems like below

我的要求是如何将模型类映射到Array，以及如何在tableList中显示它们，请帮忙

MY requirement is how to map that model classes to Array and how to display them in tableList can some one help me please

[{
  "_id" : "5470def9e0c0be27780121d7",
  "imageUrl" : "https:\/\/s3-eu-west-1.amazonaws.com\/api-static\/clubs\/5470def9e0c0be27780121d7_180.png",
  "name" : "Mondo",
  "hasVip" : false,
  "location" : {
    "city" : "Madrid"
  }
}, {
  "_id" : "540b2ff281b30f3504a1c72f",
  "imageUrl" : "https:\/\/s3-eu-west-1.amazonaws.com\/api-static\/clubs\/540b2ff281b30f3504a1c72f_180.png",
  "name" : "Teatro Kapital",
  "hasVippler" : false,
  "location" : {
    "address" : "Atocha, 125",
    "city" : "Madrid"
  }
}, {
  "_id" : "540cd44581b30f3504a1c73b",
  "imageUrl" : "https:\/\/s3-eu-west-1.amazonaws.com\/api-static\/clubs\/540cd44581b30f3504a1c73b_180.png",
  "name" : "Charada",
  "hasVippler" : false,
  "location" : {
    "address" : "La Bola, 13",
    "city" : "Madrid"
  }
}]

映射:

俱乐部:-

class Club { 

    var id: String = ""
    var name: String = ""
    var imageUrl: String = ""
    var hasVip: Bool = false
    var desc: String = ""
    var location: [Location] = []

}

位置:-

class Location {

    var country: String = ""
    var city: String = ""
    var address: String = ""
    var zip: String = ""
    var underground: [String] = []

}

NSURl会话代码:-

class BackGroundPostCall: NSObject {

    var delegate:PostProtocol?

    func callPostService(url:String,params:NSDictionary){

        print("url is===>\(url)")

        let request = NSMutableURLRequest(URL: NSURL(string:url)!)

        let session = NSURLSession.sharedSession()
        request.HTTPMethod = "POST"

        //Note : Add the corresponding "Content-Type" and "Accept" header. In this example I had used the application/json.
        request.addValue("application/json", forHTTPHeaderField: "Content-Type")
        request.addValue("application/json", forHTTPHeaderField: "Accept")

        request.HTTPBody = try! NSJSONSerialization.dataWithJSONObject(params, options: [])

        let task = session.dataTaskWithRequest(request) { data, response, error in
            guard data != nil else {
                print("no data found: \(error)")
                return
            }

            do {
                if let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as? NSArray {
                    print("Response: \(json)")
                } else {
                    let jsonStr = NSString(data: data!, encoding: NSUTF8StringEncoding)// No error thrown, but not NSDictionary
                    print("Error could not parse JSON: \(jsonStr)")
                }
            } catch let parseError {
                print(parseError)// Log the error thrown by `JSONObjectWithData`
                let jsonStr = NSString(data: data!, encoding: NSUTF8StringEncoding)
                print("Error could not parse JSON: '\(jsonStr)'")
            }
        }

        task.resume()
    }
}

推荐答案

您可以使用以下网址创建模型类: http: //www.jsoncafe.com/

You can make model class using this url : http://www.jsoncafe.com/

打开此链接，然后将json放在JSON选项卡中，然后选择所需的任何代码模板.并且如果需要，您还可以添加前缀类名和根类名，否则它是可选的.最后单击生成"按钮，您的json类就准备好了！！

open this link and put your json in JSON tab and select any Code Template that is you want. and there is you can also add prefix class name and root class name if you want other wise it is optional. and last click on generate button, your json class is ready.!!

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