Deserializing JSON when fieldnames contain spaces

Question

I'm writing a tool to read JSON files. I'm using the NewtonSoft tool to deserialize the JSOn to a C# class. Here's an example fragment:

 "name": "Fubar",
 ".NET version": "4.0",
 "binding type": "HTTP",

The field names contain spaces and other characters (the .) that are invalid in C# identifiers. What is the correct way to do this?

(Unfortunately I don't have the option of changing the JSON format.)

Answer 1

Use the JsonProperty attribute to indicate the name in the JSON. e.g.

[JsonProperty(PropertyName = "binding type")]
public string BindingType { get; set; }

Answer 2

System.Text.Json

If you're using System.Text.Json, the equivalent attribute is JsonPropertyName:

[JsonPropertyName(".net version")]
public string DotNetVersion { get; set; }

Example below:

public class Data
{
    public string Name { get; set; }

    [JsonPropertyName(".net version")]
    public string DotNetVersion { get; set; }

    [JsonPropertyName("binding type")]
    public string BindingType { get; set; }
}

// to deserialize
var data = JsonSerializer.Deserialize<Data>(json);

Answer 3

Not sure why but this did not work for me. In this example I simply return a null for "BindingType" every time. I actually found it much easier to just download the Json result as a string and then do something like:

  myString = myString.Replace(@"binding type", "BindingType")

You would do this as the step before deserializing.

Also was less text by a tad. While this worked in my example there will be situations where it may not. For instance if "binding type" was not only a field name but also a piece of data, this method would change it as well as the field name which may not be desirable.

Answer 4

If you want to initialize the Json manually, you can do:

var jsonString = "{" +
            "'name': 'Fubar'," +
            "'.NET version': '4.0'," +
            "'binding type': 'HTTP'," +
            "}";
        var json = JsonConvert.DeserializeObject(jsonString);            
        return Ok(json);

Don't forget to include using Newtonsoft.Json;