Cannot convert value of type '[String : Any]' to expected argument type 'String'

Question

I am trying to create a weather App in Swift without using Podfiles and I have reached a stumbling block in terms of passing a dictionary as an argument in parsing JSON. The location function is as follows:

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)

    }
}

So to parse the JSON I have created the following function:

private func getWeatherData(url: String, parameters: [String : String]) {
    let JsonURLString:[String: Any] = ["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()
}

The error I am having is the line Cannot convert the value of type '[String: Any]' to expected argument type 'String' in the line guard let url = URL(string: JsonURLString) else { return }. I am wondering if I have gone down a rabbit hole and maybe use an alternative approach. I wanted to use codable in Swift 4 as this is supposed to be an easier way of parsing JSON. Now in this example, I am not sure. Any help would be much appreciated.

You can't directly build a URL from a dictionary. Your issue has nothing to do with your JSON parsing code or using Codable. Your issue is creating a URL from a dictionary. — rmaddy
– rmaddy, Commented Apr 25, 2019 at 19:09
You should use URLComponents to build the URL from all of the pieces that you have. — rmaddy
– rmaddy, Commented Apr 25, 2019 at 19:14
Or should I rewrite this function and pass in the co-ordinates of "Lat" and "Long"? So for instance private func getWeatherData(Lat: Double, Long: Double) ? — AltBrian
– AltBrian, Commented Apr 25, 2019 at 19:15
The reference documentation and search. You'll find plenty of examples here. — rmaddy
– rmaddy, Commented Apr 25, 2019 at 20:05

Wide Angle Technology · Accepted Answer · 2019-04-26 07:05:28Z

1

Try this code.

JsonURLString is a [String: Any] type object not a String type. So you nee to convert it to a String type object.

private func getWeatherData(url: String, parameters: [String : String]) {
   let JsonURLString:[String: Any] = ["url": WEATHER_URL, "parameters": parameters]
   print(JsonURLString)
   let urlString = JsonURLString["url"] as? String
   guard let url = URL(string: urlString!) 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()
}

answered Apr 26, 2019 at 7:05

Wide Angle Technology

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1 Comment

AltBrian Over a year ago

Cheers Gregory will try this later.

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