我正在尝试将字典作为函数中的参数传递。我目前正在构建一个小型天气应用。我遇到的问题是尝试获取和解析JSON数据。我之前已经做过Weather应用程序,并且使用过Alamofire,但是我想不再使用podfile。 Alamofire的功能如下:
func getWeatherData(url: String, parameters: [String : String]) {
Alamofire.request(url, method: .get, parameters: parameters).responseJSON {
response in
if response.result.isSuccess {
print("Success! Got the weather data")
let weatherJSON : JSON = JSON(response.result.value!)
print(weatherJSON)
self.updateWeatherData(json: weatherJSON)
} else {
print("Error \(String(describing: response.result.error))")
self.cityLabel.text = "Connection Issues"
}
}
}
现在,我想使用可编码和解码器来访问JSON数据。我遇到的问题是设置JSONURLString变量。
private func getWeatherData(url: String, parameters: [String : String]) {
let JsonURLString:[String: String] = [url: WEATHER_URL, parameters: parameters]
print(JsonURLString)
guard let url = URL(string: JsonURLString) else { return }
URLSession.shared.dataTask(with: url) { ( data, response, err ) in
DispatchQueue.main.sync {
if let err = err {
print("Failed to get data from url:", err)
return
}
guard let data = data else { return }
do {
let decoder = JSONDecoder()
decoder.keyDecodingStrategy = .convertFromSnakeCase
let city = try decoder.decode(WeatherData.self, from: data)
self.weatherData.city = city.name
} catch {
print(error)
self.cityLabel.text = "Connection issues"
}
}
}.resume()
}
我正在使用的API来自{{3}}。下面是获取当前位置的函数。
func locationManager(_ manager: CLLocationManager, didUpdateLocations locations: [CLLocation]) {
let location = locations[locations.count - 1]
if location.horizontalAccuracy > 0 {
locationManager.startUpdatingLocation()
locationManager.delegate = nil
print("longitude = \(location.coordinate.longitude), latitude = \(location.coordinate.latitude)")
let latitude = String(location.coordinate.latitude)
let longitude = String(location.coordinate.longitude)
let params : [String : String] = ["lat" : latitude, "lon" : longitude, "appid" : APP_ID]
getWeatherData(url: WEATHER_URL, parameters: params)
}
}
问题在于上述功能无法正常工作,因为我无法从openweatherAPI解析JSON。任何帮助将不胜感激。
答案 0 :(得分:1)
let JsonURLString:[String: String]
应该设为JsonURLString:[String: Any]
,因为您的参数已经是字典。
正确的代码行应如下所示
let JsonURLString:[String: Any] = ["url": WEATHER_URL, "parameters": parameters]