How to skip property name in deserialization

Question

I'm trying to deserialize json from 3rd-party service:

{
   "route1":[
      {
         "Id":1
      }
   ],
   "route2":{
      "error":"Id not found"
   }
}

It's a Dictionary, but value can be array or an object. I need the only array, so I decided to put an empty array when I found an object in JsonConverter.

public class Item
{
    public int Id { get; set; }
}

public class JsonInfo
{
    [JsonConverter(typeof(ItemConverter))]
    public List<Item> Items{ get; set; }
}

public class ItemConverter : Newtonsoft.Json.JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        switch (reader.TokenType)
        {
            case JsonToken.String:
            case JsonToken.StartObject:
                return new List<Item>();

            case JsonToken.StartArray:
                var array = JArray.Load(reader);
                var data = array.Select(ParseValue).ToList();
                return data;

            default:
                return null;
        }
    }

    public override bool CanConvert(Type objectType)
    {
        return false;
    }

    private Item ParseValue(JToken value)
    {
        switch (value.Type)
        {
            case JTokenType.Object:
                return value.ToObject<Item>();

            default:
                return null;
        }
    }
}

But when I'm trying to deserialize Dictionary<string, JsonInfo> it raises an error (must be json array). I think the problem that converter trying to find JsonInfo property in json, instead of array inside this class.

Maybe I missed something?
Do we have an attribute, that allows skipping property name?

Sure [there is](https://stackoverflow.com/q/10169648/1997232) an attribute. Problem solved? — Sinatr, Jun 29 '21 at 07:44
Ingore doesn't solve it, because I need to deserialize this property. I do not need to deserialize the JsonInfo object, I need to deserialize the list inside this class, i.e. skip object, but at the same time I can't use Dictionary>, because I need to apply custom JsonConverter — Aborigen, Jun 29 '21 at 10:22

score 0 · Answer 1 · answered Jun 29 '21 at 11:08

You don't need to create a custom converter.

You can achieve the same with following few lines of code:

var semiParsedJson = JObject.Parse(rawJson);
var result = new Dictionary<string, List<Item>>(); 
foreach (var item in semiParsedJson)
    if (item.Value is JArray)
        result.Add(item.Key, item.Value.ToObject<List<Item>>());
    else 
        result.Add(item.Key, new List<Item>>());

We semi parse the json to a JObject
We are iterating through its properties via a simple foreach where item's type is KeyValuePair<string, JToken?>
If the Value is a JArray (not a JObject) then we simply populate the dictionary with the content of Value
Finally we let Newtonsoft to convert the Value to a List<Item> collection on our behalf

With this approach you don't need to write a custom converter and you don't need to decorate your domain model with serialization attributes.

Thank you for your answer. It seems to be working solution, but I don't want to change default deserializer. (it will affect other endpoints) — Aborigen, Jun 29 '21 at 11:28
You don't have to change the default deserializer. You can simply put this code into a method and call it after you have read the 3rd party service's response. — Peter Csala, Jun 29 '21 at 11:56

score 0 · Accepted Answer · edited Jun 29 '21 at 11:54

I solved my issue by applying custom converter not to the property, but to the entire class

[JsonConverter(typeof(ItemConverter))]
public class JsonInfo
{
    public List<Item> Items{ get; set; }
}

public class ItemConverter: JsonConverter
{
    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        switch (reader.TokenType)
        {
            case JsonToken.String:
                return null;
            case JsonToken.StartObject:
                reader.Skip();
                return new JsonInfo{ Items = new List<Item>() };

            case JsonToken.StartArray:
                var array = JArray.Load(reader);
                var data = array.Select(ParseValue).ToList();
                return new JsonInfo{ Items = data };

            default:
                return null;
        }
    }

    public override bool CanConvert(Type objectType)
    {
        return false;
    }

    private Item ParseValue(JToken value)
    {
        switch (value.Type)
        {
            case JTokenType.Object:
                return value.ToObject<Item>();

            default:
                return null;
        }
    }
}

How to skip property name in deserialization

2 Answers2