Question

有人可以告诉我如何将这些披萨店的名称打印出来吗？我的应用程序打印出预期的"Status Code: 200"。但是，我的控制台只显示空括号[]。我怀疑我没有正确地从我的JSON对象中提取值。

我已将此链接用于我的API。

问题

如何从序列化的JSON对象中正确获取值？

相关代码：

    // Response
    if let httpResponse = response as? NSHTTPURLResponse where httpResponse.statusCode == 200, let data = data {
        print("Status Code: \(httpResponse.statusCode)")



        do {
            let json = try NSJSONSerialization.JSONObjectWithData(data, options: .MutableContainers)

            if let pizzaPlaces = json["response"] as? [[String: AnyObject]] {
                for place in pizzaPlaces {
                    if let name = place ["name"] as? String {
                        self.PizzaClass.append(name)
                    }
                }
            }
        } catch {
            print("Error Serializing JSON Data: \(error)")
        }
        print(self.PizzaClass)


    }
}).resume()

Answer 1

您需要将NSJSONSerialization.JSONObjectWithData结果转换为[String:AnyObject]。

    let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]

一旦你拥有了所有你需要做的就是注意你正在施展的东西。以下面的代码为例。如果我们想使用response获取jsonObject["response"]对象，我们会使用哪种数据结构？

"response": {
        "venues": [{
            //... continues
        }]
    }

在左侧，我们有"response"这是一个字符串，在右边我们有{}，这是一个AnyObject。所以我们有[String: AnyObject]。你只需要考虑你一件一件地处理什么对象。下面是一个可以粘贴到您的应用程序中的工作示例。

完整的工作代码：

func getJson() {

    let request = NSMutableURLRequest(URL: NSURL(string: "https://api.foursquare.com/v2/venues/search?client_id=0F5M0EYOOFYLBXUOKTFKL5JBRZQHAQF4HEM1AG5FDX5ABRME&client_secret=FCEG5DWOASDDYII4U3AAO4DQL2O3TCN3NRZBKK01GFMVB21G&v=20130815%20&ll=29.5961,-104.2243&query=burritos")!)
    let session = NSURLSession.sharedSession()

    request.HTTPMethod = "GET"
    request.addValue("application/json", forHTTPHeaderField: "Accept")

    let task = session.dataTaskWithRequest(request) { (data: NSData?, response: NSURLResponse?, error: NSError?) -> Void in

        guard let testResponse = response as? NSHTTPURLResponse else {
            print("\(response)")
            return
        }

        guard let status = HTTPStatusCodes(rawValue: testResponse.statusCode) else {
            print("failed to unwrap status")
            return
        }
        print(status)

        switch status {
        case .Created:
            print("ehem")
        case .BadRequest:
            print("bad request")
        case .Ok:
            print("ok")

            guard let returnedData = data else {
                print("no data was returned")
                break
            }
            do {
                let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]

                guard let response = jsonObject["response"] as? [String: AnyObject] else { return }     
                guard let venues = response["venues"] as? [AnyObject] else { return }
                guard let location = venues[0]["location"] as? [String:AnyObject] else { return }
                guard let formattedAddress = location["formattedAddress"] else { return }

                print("response: \n\n \(response)\n------")
                print("venues : \n\n \(venues)\n-------")
                print("location : \n\n \(location)\n------")
                print("formatted address : \n \(formattedAddress)")

            } catch let error {
                print(error)
            }

            // update user interface
            dispatch_sync(dispatch_get_main_queue()) {
                print("update your interface on the main thread")
            }
        }  
    }
    task.resume()
}

将它放在我们自己的文件中，我们在类声明之外，

  enum HTTPStatusCodes : Int {
        case Created = 202
        case Ok = 200
        case BadRequest = 404 
    }

Answer 2

这不是你想要的，但是因为你是Swift的新手，请看看Alamofire。它为您处理JSON序列化。当你需要连锁电话时，PromiseKit是超级光滑的。

         Alamofire.request(.GET, url).responseJSON {response in

            switch (response.result) {
            case .Success(let value):

                let pizzas = JSON(value).arrayValue

                for place in pizzaPlaces {
                   if let name = place ["name"] as? String {
                     self.PizzaClass.append(name)
                   }
                }                 

            case .Failure(let error):
                if let data = response.data, let dataString = String(data: data, encoding: NSUTF8StringEncoding) {
                    print("ERROR data: \(dataString)")
                }
                print("ERROR: \(error)")
            }
        }

如何从复杂的JSON对象中获取值？

问题

2 个答案: