Unable To Find A Constructor

Question

ScreenShot Of Error i have a json file which contains some data and i want to parse it but every time when i want to Deserialize Object it give me this error "Unable to find a constructor. A Class should either have a default constructor, one constructor with arguments or a constructor marked with json constructor i dont know what to do because i did not work with json before it is my first time working with json. here is my json :

{
   "Addition":
   {
       "Easy": [

                "New York Bulls",
                "Los Angeles Kings",
                "Golden State Warriros",
                "Huston Rocket"
            ],
      "Medium":[
                "New York Bulls",
                "Los Angeles Kings",
                "Golden State Warriros",
                "Huston Rocket"
      ],
       "Difficult": [
               "New York Bulls",
               "Los Angeles Kings",
               "Golden State Warriros",
               "Huston Rocket"
      ] 
   }
}

here is my model class

public class Addition
{

     public List<string> Easy { get; set; }
        public List<string> Medium { get; set; }
        public List<string> Difficult { get; set; }

    public Addition() { }

}

here is my function in which i am Deserialize object

private void ReadJson()
    {
        var assembly = typeof(WordProblemsScreen).GetTypeInfo().Assembly;
        Stream stream = 
assembly.GetManifestResourceStream("MathRandomizer.demo.json");

        using (var reader = new System.IO.StreamReader(stream))
        {

            string json = reader.ReadToEnd();

            JObject jObject = JObject.Parse(json);
            JToken jUser = jObject["Addition"];


            var addition = JsonConvert.DeserializeObject<Addition>(json);




        }
    }

I think you don´t even need this constructor and can rely on the default-constructor instead. `JsonConvert` will use that if no other one is provided, and set the properties appropriately. — MakePeaceGreatAgain, Apr 12 '19 at 06:53
@HimBromBeere i have tried what you are saying but the error was still the same — Masti Broz Top, Apr 12 '19 at 06:55
I think you have another `Addition` class. Click in the middle of `` and then press `[F12]`. See where that takes you. — H H, Apr 12 '19 at 07:05
`var addition = JsonConvert.DeserializeObject(json);` <-- you're trying to deserialize your JSON into the wrong object. You need a root object. — ProgrammingLlama, Apr 12 '19 at 07:06
@HenkHolterman no i am using the right class it takes me to the class the i defined in the question above — Masti Broz Top, Apr 12 '19 at 07:08
@John can u please explain bit more or write some valid code — Masti Broz Top, Apr 12 '19 at 07:09
@Masti Your C# classes have to match your JSON objects. The first `{` in you JSON is the object you need to deserialize into. The root object of your JSON holds your `Additions` class - you seem to be aware of this from the `JToken jUser = jObject["Addition"];` line. — ProgrammingLlama, Apr 12 '19 at 07:11
Always post the exact error message(s), with line numbers and stack trace if possible. — H H, Apr 12 '19 at 07:11
Possible duplicate of [Deserialize JSON to C# Classes](https://stackoverflow.com/questions/25052293/deserialize-json-to-c-sharp-classes) — ProgrammingLlama, Apr 12 '19 at 07:11
I've reverted your edit because it invalidates the answers already provided. — ProgrammingLlama, Apr 12 '19 at 07:28
[It works for me](https://rextester.com/LTP54588) once I adjust your deserialize call to reflect that `RootObject` is nested in `Addition` (for some reason?). As I said before: the only logical conclusion now is that your JSON is different. Without knowing what your JSON really is, we have no way of helping you. I've printed the JSON in the linked fiddle - the lack of newlines doesn't make a difference. Compare it to your actual JSON by also doing `Console.WriteLine(json)` in your own code. — ProgrammingLlama, Apr 12 '19 at 07:32
I also don't believe you're getting the same "Unable to find a constructor." error. — ProgrammingLlama, Apr 12 '19 at 07:34
i am still getting this error can we do some private chat so i can show you that i am still getting this error — Masti Broz Top, Apr 12 '19 at 07:36

score 2 · Answer 1 · answered Apr 12 '19 at 07:09

2

Paste your Json as C# using Edit - Paste Special - Paste JSON as C# or use one of the online JSON to C# converters and you will see the issue:

public class Addition
{
    public List<string> Easy { get; set; }
    public List<string> Medium { get; set; }
    public List<string> Difficult { get; set; }
}

public class RootObject
{ 
   public Addition Addition { get; set; }
}

answered Apr 12 '19 at 07:09

Ian Mercer

38,490
8
97
133

@Masti Deserialize to RootObject. – ProgrammingLlama Apr 12 '19 at 07:26

ProgrammingLlama · Accepted Answer · 2019-04-12T07:16:21.057

0

You need a root object:

public class RootObject
{
    public Addition Addition { get; set; }
}

var result = JsonConvert.DeserializeObject<RootObject>(json);
Console.WriteLine(result.Addition.Easy.First());

Try it online

And it even works with your old code with the [JsonConstructor] attribute.

Simply put: your C# classes must match the JSON you hope to deserialize.

Imagine you have the following JSON:

{
    "person": {
        "name": "John"
    }
}

You cannot simply deserialize it into the following class:

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

Why? Because the "person" object is not your root object. The above class serialized as JSON simply looks like this:

{
    "name": "John"
}

So to deserialize the first sample above, we need a root object:

public class RootObject
{
    public Person Person { get; set; }
}

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

And now we can deserialize that JSON (the one that contains the "person" key) to RootObject

JSON is really simple - things just have to match.

edited Apr 12 '19 at 07:16

answered Apr 12 '19 at 07:09

ProgrammingLlama

36,677
7
67
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i am getting the same error this is what i have done i made a new class name RootObject and put the above code in that class then i put the last line in my readjson function but still the same error – Masti Broz Top Apr 12 '19 at 07:21
OK how come we don't get the error in the two example fiddles? You're doing something differently. – ProgrammingLlama Apr 12 '19 at 07:22
i have a separate json file from where i want to read values but you are writing json in class – Masti Broz Top Apr 12 '19 at 07:24
i have copy and pasted same your code but still the same error – Masti Broz Top Apr 12 '19 at 10:28
@Masti please upload a sample that reproduces the issue to GitHub and I'll take a look. – ProgrammingLlama Apr 12 '19 at 10:49
hey john i have tested your code in a new separate project and its working but when i use your code in my existing project it gives me same error even i make a new content page but still same issue in my existing project – Masti Broz Top Apr 15 '19 at 06:57
@Masti Check the version of JSON.NET you're using. Sometimes there are minor discrepancies between the versions. It's best to use the latest, I think. Also check that you don't have two classes (different to two files) with the same name, but in different namespaces. It might be that you are deserializing to the wrong class. – ProgrammingLlama Apr 15 '19 at 06:59
version is 12.0.1 and i have only one class – Masti Broz Top Apr 15 '19 at 07:05
@Masti No idea then I'm afraid. There's something strange about your project. – ProgrammingLlama Apr 15 '19 at 07:06
i am also feeling strange because same code is running 100% right on separate project but not in my project even now i dont have any duplicate classes etc etc – Masti Broz Top Apr 15 '19 at 07:09
Double check that by clicking on "Addition" in your deserialization code and then press F12 on your keyboard. It will take you to the definition. Is it the class you expect to see? – ProgrammingLlama Apr 15 '19 at 07:10
check my question again i have added a screenshot of the error – Masti Broz Top Apr 15 '19 at 07:15
That's _very strange_. I've never had to explicitly declare a constructor. Try adding `public RootObject() { }` as a default parameterless constructor. This isn't exactly a solution if it does work, since JSON.NET shouldn't need you to explicitly declare it. – ProgrammingLlama Apr 15 '19 at 07:17
same issue bro btw this code is running on separate project – Masti Broz Top Apr 15 '19 at 07:20
Can you try downgrading JSON.NET one major version, just in case it's a bug. – ProgrammingLlama Apr 15 '19 at 07:20
but bro then why its running fine on separate project like the versions are same on both projects 12.0.1 – Masti Broz Top Apr 15 '19 at 07:23
Good point. I honestly have no clue at this stage. Without seeing the entire failing project and being able to test it, I can't really advise. – ProgrammingLlama Apr 15 '19 at 07:23
[Related?](https://stackoverflow.com/questions/27938529/jsonserializationexception-unable-to-find-a-constructor-on-xamarin-android) – ProgrammingLlama Apr 15 '19 at 07:24
i have downgraded but same issue – Masti Broz Top Apr 15 '19 at 07:29
1

thanks man i solve it by going in to android options and changing linking to sdk assemblies only – Masti Broz Top Apr 15 '19 at 07:36

score 0 · Answer 3 · answered Apr 12 '19 at 07:18

The exception occurs because you try to convert your JSON content into the wrong object. As your JSON contains an attribute "Addition" in top, it is considered as a property of some parent class. So it expects some class like this:

namespace JsonStuff
{
    public class SomeParentClass
    {
        public Addition AdditionObject { get; set; }

        public SomeParentClass()
        {

        }
    }
}

If you try it the other way serializing an object from type "Addition" and have a look on the outcoming json string, you will see the difference:

var filledObject = new Addition()
{
    Easy = new List<string>()
    {
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    },
    Medium = new List<string>()
    {
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    },
    Difficult = new List<string>()
    {
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    }
};

var serializedObject = JsonConvert.SerializeObject(filledObject);

The result will be:

{
    "Easy":
    [
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    ],
    "Medium":
    [
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    ],
    "Difficult":
    [
        "New York Bulls",
        "Los Angeles Kings",
        "Golden State Warriros",
        "Huston Rocket"
    ]
}

and there you see the difference: There is no "Addition" property on top.

You can either create a parent class like the one I posted on top or you change your json File to the format like the example at the bottom. — rekcul, Apr 12 '19 at 07:27

Unable To Find A Constructor

3 Answers3