JsonSerializer deserialize to anonymous type-CodePudding

How do I deserialize a JSON to an anonymous type?

I have this working code that's:

defining a JSON with Foo and Bar fields
defining a model as anonymous type with Foo property
deserializing the JSON using the generic method

    [Fact]
    public void Test1()
    {
        const string json = """
        {
            "Foo": "a",
            "Bar": "b"
        }
        """;

        var model = new
        {
            Foo = default(string),
        };

        var deserialized = Deserialize(json, model);

        var a = deserialized.Foo;

        a.Should().Be("a");
    }

    private T Deserialize<T>(string json, T model) => JsonSerializer.Deserialize<T>(json) ?? model;

I would like to do the same without the generic Deserialize method. Below the code:

    [Fact]
    public void Test2()
    {
        const string json = """
        {
            "Foo": "a",
            "Bar": "b"
        }
        """;

        var model = new
        {
            Foo = default(string),
        };

        var deserialized = JsonSerializer.Deserialize<??????>(json);
        string a = deserialized.Foo;

        a.Should().Be("a");
    }

Not sure what to put instead of the ??????. Ideally I'd like to specify something like typeof(model) or model.GetType(), but those are not accepted by the compiler

CodePudding user response：

Not sure that this will suit your actual use case, but for current one you can try using Dictionary:

var deserialized = JsonSerializer.Deserialize<Dictionary<string, string>>(json);
string a = deserialized["Foo"];

Or just introduce the class, with records (C# 9) and file keyword (C# 11) it should extremely simple and effortless:

file record RecordForTest1(string Foo, string Bar);

[Fact]
public void Test1()
{
    // ...
    var deserialized = JsonSerializer.Deserialize<RecordForTest1>(json);
}

CodePudding user response：

You can use ExpandoObject.

var deserialized = JsonSerializer.Deserialize<ExpandoObject>(json);
string a = deserialized.Foo.ToString();
...

If you want you can also create extension method like below:

public static dynamic? DeserializeAsDynamic(this string json)
{
    var result = JsonConvert.DeserializeObject<ExpandoObject>(json);
    return result;
}

Sample usage:

dynamic obj = json.DeserializeAsDynamic();