Question

if let jsonObj = jsonObj as? [String: Any],

let weatherDictionary = jsonObj["weather"] as? [String: Any],

let weather = weatherDictionary["description", default: "clear sky"] as? 
NSDictionary {

                            print("weather")

                        DispatchQueue.main.async {
                            self.conditionsLabel.text = "\(weather)"
                        }
                    }

// to display weather conditions in "name" from Open Weather

"weather":[{"id":800,"main":"Clear","description":"clear sky","icon":"01n"}]

//No errors, but code is not printing or displaying in App.

Answer 1

除非您能为上下文提供更多代码，否则我不确定如何帮助解决您的确切问题。然而，

您可以尝试使用Swift 4附带的内置解码。检查出here.基本上，您创建了一个为响应对象建模的类，如下所示：

struct Weather: Decodable {
    var id: Int
    var main: String
    var description: String
    var icon: String
}

然后解码它：

let decoder = JSONDecoder()
let weather = try decoder.decode(Weather.self, from: jsonObj)

它会神奇地解码为您需要的数据！如果这不起作用，请告诉我，如果您有更多代码上下文可供我帮助，请发表评论。

Answer 2

我在此处放置了完整的演示，以展示如何发送HTTP请求并解析JSON响应。

注意：如果您使用HTTP请求而不是HTTPS请求，请配置ATS。

演示网址为＆＃34; http://samples.openweathermap.org/data/2.5/forecast?q=M%C3%BCnchen,DE&appid=b6907d289e10d714a6e88b30761fae22＆＃34;。
JSON格式如下，演示了如何获取城市名称。

{ cod: "200", message: 0.0032, cnt: 36, list: [...], city: { id: 6940463, name: "Altstadt", coord: { lat: 48.137, lon: 11.5752 }, country: "none" } }

完整演示如下。它显示了如何使用URLSessionDataTask和JSONSerialization。

class WeatherManager {
  static func sendRequest() {
    guard let url = URL(string: "http://samples.openweathermap.org/data/2.5/forecast?q=M%C3%BCnchen,DE&appid=b6907d289e10d714a6e88b30761fae22") else {
      return
    }
    // init dataTask
    let dataTask = URLSession.shared.dataTask(with: url) { (data, response, error) in
      let name = WeatherManager.cityName(fromWeatherData: data)
      print(name ?? "")
    }
    // send the request
    dataTask.resume()
  }

  private static func cityName(fromWeatherData data: Data?) -> String? {
    guard let data = data else {
      print("data is nil")
      return nil
    }

    do {
      // convert Data to JSON object
      let jsonObject = try JSONSerialization.jsonObject(with: data, options: [])
      print(jsonObject)
      if let jsonObject = jsonObject as? [String: Any],
        let cityDic = jsonObject["city"] as? [String: Any],
        let name = cityDic["name"] as? String {
        return name
      } else {
        return nil
      }
    } catch {
      print("failed to get json object")
      return nil
    }
  }
}

Swift没有在天气API中打印或显示应用程序中的名称？

2 个答案: