how to use JsonConvert.DeserializeObject for values without name

Question

I have a json string like this: {"status":false,"data":{"errors":{"":"error45"}}}

I cant make a class for this json string, because the last part has no name => ""

I test this class:

public class Result
{
    public bool status { set; get; }
    public ResultDetail data { set; get; }
}

public class ResultDetail
{
    public ErrorDetails errors { set; get; }       
}

public class ErrorDetails
{        
    public string abc { set; get; }
}

but abc returns null !!!!

Add this to `abc` -> `[JsonProperty("")]` – Trevor May 21 '21 at 20:43 — Trevor, May 21 '21 at 20:43
I dont think that is valid json – Jonathan Alfaro May 21 '21 at 21:35 — Jonathan Alfaro, May 21 '21 at 21:35

Jawad · Accepted Answer · 2021-05-21T20:51:12.880

1

You can use the Dictionary<string, string> for the errors.

public class Result
{
    public bool status { set; get; }
    public ResultDetail data { set; get; }
}

public class ResultDetail
{
    public Dictionary<string, string> errors { set; get; }
}

and use the following to deserialize and access the dictionary.

var result = JsonConvert.DeserializeObject<Result>(json);
Console.WriteLine(result.data.errors.Values.FirstOrDefault());

Or you can assign the value to a variable

obj.data.errors.TryGetValue("", out string error);
Console.WriteLine(error);

I will mention this as well that the property names should be Proper names (first letter Capital). To be able to deserialize properly, use the [JsonProperty("corresponding_json_key")] to each of your properties of your classes to conform to C# standards.

public class Result
{
    [JsonProperty("status")]
    public bool Status { set; get; }

    [JsonProperty("data")]
    public ResultDetail Data { set; get; }
}

edited May 21 '21 at 20:51

answered May 21 '21 at 20:41

Jawad

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Thanks for your Answer! Im beginner in C#! when I wrote [JsonProperty("data")], I got the error: "the type or namespace jsonproperty couldnt be found! or when I wrote Dictionary, I got the same error for it. Please help me, what should I do? – Alireza Karimian May 22 '21 at 15:47
You need to include `using Newtonsoft.Json` directive on top. Make sure you have newtonsoft json package installed as well – Jawad May 22 '21 at 16:12
Thanks for your comment. It was helpful – Alireza Karimian May 27 '21 at 05:46
how can I use Dictionary instruction? I have error in using it, yet! – Alireza Karimian May 27 '21 at 05:49

how to use JsonConvert.DeserializeObject for values without name

1 Answers1