JSON Not Deserializing certain values

Question

I am consuming an API which returns the following JSON

{"coord":{"lon":-1.38,"lat":54.91},"weather":[{"id":801,"main":"Clouds","description":"few clouds","icon":"02d"}],"base":"stations","main":{"temp":285.83,"pressure":1013,"humidity":76,"temp_min":285.15,"temp_max":286.15},"visibility":10000,"wind":{"speed":5.1,"deg":240},"clouds":{"all":20},"dt":1536913200,"sys":{"type":1,"id":5104,"message":0.0049,"country":"GB","sunrise":1536903391,"sunset":1536949457},"id":2636531,"name":"Sunderland","cod":200}

I call the API and return the JSON as a string (strJson), from strJson I use the following code to deserialize the JSON and return the values.

JsonResponse res = JsonConvert.DeserializeObject<JsonResponse>(strJson);

I am able to return the items that reside within "weather" by using the following inside my JsonResponse class (I'm only returning "main" at the moment, which works successfully;

class JsonResponse
{

    public object coord { get; set; }
    public IList<Weather> weather { get; set; }
}

}

 class Weather
    {
    public string main { get; set; }
    }

I have tried all sorts to return the data that resides within "coord" and "sys" but i keep getting the error "Cannot deserialize the JSON array (e.g. [1,2,3]) into type ' ' because type requires JSON object (e.g. {“name”:“value”}) to deserialize correctly".

I am able to return into into an "object" using

public object coord {get; set;}

however this isn't effective as I'm still unable to quickly extract the variables lon & Lat as well as their values.

Any help would be appreciated as I've asked a similar question before, however the answers I've found stop me from being able to extract the values from "weather" so I'm stumped.

Rafalon · Accepted Answer · 2018-09-14 10:11:55Z

3

What's wrong with this?

class JsonResponse
{
    public CoordObject coord {get;set;}
}

class CoordObject
{
    public double lon {get;set;}
    public double lat {get;set;}
}

You should then be able to access res.coord.lon and res.coord.lat.

With a quick search "json detect class", I found this helpful site: json2csharp which generated the following classes for your JSON:

public class Coord
{
    public double lon { get; set; }
    public double lat { get; set; }
}

public class Weather
{
    public int id { get; set; }
    public string main { get; set; }
    public string description { get; set; }
    public string icon { get; set; }
}

public class Main
{
    public double temp { get; set; }
    public int pressure { get; set; }
    public int humidity { get; set; }
    public double temp_min { get; set; }
    public double temp_max { get; set; }
}

public class Wind
{
    public double speed { get; set; }
    public int deg { get; set; }
}

public class Clouds
{
    public int all { get; set; }
}

public class Sys
{
    public int type { get; set; }
    public int id { get; set; }
    public double message { get; set; }
    public string country { get; set; }
    public int sunrise { get; set; }
    public int sunset { get; set; }
}

public class RootObject
{
    public Coord coord { get; set; }
    public List<Weather> weather { get; set; }
    public string @base { get; set; }
    public Main main { get; set; }
    public int visibility { get; set; }
    public Wind wind { get; set; }
    public Clouds clouds { get; set; }
    public int dt { get; set; }
    public Sys sys { get; set; }
    public int id { get; set; }
    public string name { get; set; }
    public int cod { get; set; }
}

edited Sep 14, 2018 at 10:11

answered Sep 14, 2018 at 10:01

Rafalon

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2 Comments

Rafalon Over a year ago

Glad it helped :)

Pac0 Over a year ago

Yes, it might be some times surprising, but having "nothing" deserialized (instead of an Exception) if the JSON doesn't match the correct class definition is the intended default behavior.

user5292841 · Accepted Answer · 2018-09-14 10:12:38Z

0

You need to use dynamic for the object see here: Deserialize json object into dynamic object using Json.net

I can give a code example if you like, but there is one on the page.

answered Sep 14, 2018 at 10:12

user5292841

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