How to get input parameter Type using GetMethod?

Question

Good Afternoon Everyone, I am trying to dynamically invoke a function by passing its appropriate parameters. Let's say the function looks like this:

public string CreatePerson(Person p)

Object p is received as Json and I want to deserialize it into the appropriate runtime Type depending on the parameter Type so that I can pass it into the Newtonsoft.Json library function JsonConvert.DeserializeObject (jsonReceived).

Below is my code:

m = this.GetType().GetMethod(method);
List<object> a = new List<object>();
foreach (var param in m.GetParameters())
{
    //have to convert args parameter to appropriate function input
     a.Add(ProcessProperty(param.Name, param.ParameterType, args));

 }

 object invokeResult = null;
 invokeResult = m.Invoke(this, a.ToArray());


private object ProcessProperty(string propertyName, Type propertyType, string    jsonStringObject)
 {
     if (propertyType.IsClass && !propertyType.Equals(typeof(String)))
      {
          var argumentObject = Activator.CreateInstance(propertyType);
          argumentObject = JsonConvert.DeserializeObject<propertyType>(jsonStringObject);
           return argumentObject;
      }
  }

I get the following error :

The type or namespace name 'propertyType' could not be found (are you missing a using directive or an assembly reference?)

Where am I approaching this wrong? How can I get the parameter Type dynamically during runtime so it can handle types other than Person and be able to pass it to DeserializeObject?

Why not just use JsonConvert.DeserializeObject(jsonStringObject, propertyType) — dbc
– dbc, Commented Dec 29, 2014 at 20:23

Chris · Accepted Answer · 2014-12-29 20:23:39Z

4

The problem is that generics are done at compile time and you only know the type at runtime. Essentially the compiler thinks that propertyType should be a compiled type and not a variable of type Type.

Fortunately there are overloads that will let you do what you want such as DeserializeObject(String, Type)

Use like this:

argumentObject = JsonConvert.DeserializeObject(jsonStringObject, propertyType);

answered Dec 29, 2014 at 20:23

Chris

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1 Comment

Chris Over a year ago

No problem. You will find that quite often anything generic along these lines has an overload like this. And hopefully you understand a bit more about generics now so you can see why this problem happens and know that you should look for alternatives (like other overloads) if you come across something similar. :)

vcsjones · Accepted Answer · 2014-12-29 20:23:21Z

3

You cannot use a runtime System.Type propertyType as a type parameter to a generic method. Instead, use the overload of DeserializeObject that takes a runtime type:

argumentObject = JsonConvert.DeserializeObject(jsonStringObject, propertyType);

answered Dec 29, 2014 at 20:23

vcsjones

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