tl:dr JsonNode is the recommended way but dynamic typing with deserializing to ExpandoObject works and I am not sure why.

It is not possible to deserialize to dynamic in the way you want to. JsonSerializer.Deserialize<T>() casts the result of parsing to T. Casting something to dynamic is similar to casting to object

Type dynamic behaves like type object in most circumstances. In particular, any non-null expression can be converted to the dynamic type. The dynamic type differs from object in that operations that contain expressions of type dynamic are not resolved or type checked by the compiler. The compiler packages together information about the operation, and that information is later used to evaluate the operation at run time

docs.

The following code snippet shows this happening with your example.

var jsonString = "{\"foo\": \"bar\"}";
dynamic data = JsonSerializer.Deserialize<dynamic>(jsonString);
Console.WriteLine(data.GetType());

Outputs: System.Text.Json.JsonElement

The recommended approach is to use the new JsonNode which has easy methods for getting values. It works like this:

JsonNode data2 = JsonSerializer.Deserialize<JsonNode>(jsonString);
Console.WriteLine(data2["foo"].GetValue<string>());

And finally trying out this worked for me and gives you want you want but I am struggling to find documentation on why it works because according to this issue it should not be supported but this works for me. My System.Text.Json package is version 4.7.2

dynamic data = JsonSerializer.Deserialize<ExpandoObject>(jsonString);
Console.WriteLine(data.GetType());
Console.WriteLine(data.foo);

@NALB, Welcome to Microsoft Q&A， based on my test, I reproduced the same problem with you. The first error due to the url in the xml can not be identified. The second error means that we could not change the json name programmatically. I suggest that you could get the name first and replace the string in the end. Here is a code example you could refer to.

            var json = File.ReadAllText("B:\\ｔ.ｊｓｏｎ");
            var jsonReturn = JsonConvert.DeserializeObject(json);
            string pagestr = string.Empty;
            string str = "";
            foreach (var item in jsonReturn.pages)
            {
                pagestr += item.Name;
                str = Regex.Replace(item.Name, "://", "_");
                str = Regex.Replace(str, ":", "_");
            }
            json=json.Replace(pagestr,str);
            var data = "{\"Row\": " + json + "}";
            XmlDocument doc = JsonConvert.DeserializeXmlNode(data.ToString(), "root");
            Console.WriteLine( doc.InnerXml);

Tested result:

Hope my advice could help you. Best Regards, Jack

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Check one of approaches:

public class Data
{
    public string @class { get; set; }
    public Value value { get; set; }
}

public class Value
{
    public string[] groupNames { get; set; }
    public string itemType { get; set; }
    public string[] tags { get; set; }
    public string label { get; set; }
    public string category { get; set; }
}

. . .

Dictionary dictionary = JsonSerializer.Deserialize>( json );

where json is your string. See also other parameters of Deserialize.

Sample usage:

string category = dictionary["TreppenhausOG"].value.category;

If required, use "Manage NuGet Packages" window to add a reference to System.Text.Json.

you can deserialize to a Dictionary:

var dict = System.Text.Json.JsonSerializer.Deserialize<Dictionary<string, string>>(json)

Or just deserialize to Object which will yield a JsonElement that you can call GetProperty on.

How about something like this?

var myResponseClass = new MyResponseClass();

dynamic myClass = JsonSerializer.Deserialize<ExpandoObject>("{\"fixedProperty\":\"Hello\",\"dynamicProperty\": {\"attributeOne\":\"One\",\"attributeTwo\":\"Two\"}}");
dynamic myProperty = JsonSerializer.Deserialize<ExpandoObject>(myClass.dynamicProperty.ToString());

myResponseClass.FixedProperty = myClass.fixedProperty.ToString();
myResponseClass.DynamicProperty = myProperty;