Dynamically Add C# Properties at Runtime

C#DynamicReflectionProperties

C# Problem Overview


I know there are some questions that address this, but the answers usually follow along the lines of recommending a Dictionary or Collection of parameters, which doesn't work in my situation.

I am using a library that works through reflection to do lots of clever things with objects with properties. This works with defined classes, as well as dynamic classes. I need to take this one step further and do something along these lines:

public static object GetDynamicObject(Dictionary<string,object> properties) {
    var myObject = new object();
    foreach (var property in properties) {
        //This next line obviously doesn't work... 
        myObject.AddProperty(property.Key,property.Value);
    }
    return myObject;
}

public void Main() {
    var properties = new Dictionary<string,object>();
    properties.Add("Property1",aCustomClassInstance);
    properties.Add("Property2","TestString2");
    
    var myObject = GetDynamicObject(properties);
    
    //Then use them like this (or rather the plug in uses them through reflection)
    var customClass = myObject.Property1;
    var myString = myObject.Property2;
   
}

The library works fine with a dynamic variable type, with properties assigned manually. However I don't know how many or what properties will be added beforehand.

C# Solutions


Solution 1 - C#

Have you taken a look at ExpandoObject?

see: https://docs.microsoft.com/en-us/dotnet/api/system.dynamic.expandoobject

From MSDN:

> The ExpandoObject class enables you to add and delete members of its instances at run time and also to set and get values of these members. This class supports dynamic binding, which enables you to use standard syntax like sampleObject.sampleMember instead of more complex syntax like sampleObject.GetAttribute("sampleMember").

Allowing you to do cool things like:

dynamic dynObject = new ExpandoObject();
dynObject.SomeDynamicProperty = "Hello!";
dynObject.SomeDynamicAction = (msg) =>
	{
		Console.WriteLine(msg);
	};

dynObject.SomeDynamicAction(dynObject.SomeDynamicProperty);

Based on your actual code you may be more interested in:

public static dynamic GetDynamicObject(Dictionary<string, object> properties)
{
	return new MyDynObject(properties);
}

public sealed class MyDynObject : DynamicObject
{
	private readonly Dictionary<string, object> _properties;

	public MyDynObject(Dictionary<string, object> properties)
	{
		_properties = properties;
	}

	public override IEnumerable<string> GetDynamicMemberNames()
	{
		return _properties.Keys;
	}

	public override bool TryGetMember(GetMemberBinder binder, out object result)
	{
		if (_properties.ContainsKey(binder.Name))
		{
			result = _properties[binder.Name];
			return true;
		}
		else
		{
			result = null;
			return false;
		}
	}

	public override bool TrySetMember(SetMemberBinder binder, object value)
	{
		if (_properties.ContainsKey(binder.Name))
		{
			_properties[binder.Name] = value;
			return true;
		}
		else
		{
			return false;
		}
	}
}

That way you just need:

var dyn = GetDynamicObject(new Dictionary<string, object>()
	{
		{"prop1", 12},
	});

Console.WriteLine(dyn.prop1);
dyn.prop1 = 150;

Deriving from DynamicObject allows you to come up with your own strategy for handling these dynamic member requests, beware there be monsters here: the compiler will not be able to verify a lot of your dynamic calls and you won't get intellisense, so just keep that in mind.

Solution 2 - C#

Thanks @Clint for the great answer:

Just wanted to highlight how easy it was to solve this using the Expando Object:

var dynamicObject = new ExpandoObject() as IDictionary<string, Object>;
foreach (var property in properties) {
	dynamicObject.Add(property.Key,property.Value);
}      

Solution 3 - C#

you could deserialize your json string into a dictionary and then add new properties then serialize it.

var jsonString = @"{}";

var jsonDoc = JsonSerializer.Deserialize<Dictionary<string, object>>(jsonString);

jsonDoc.Add("Name", "Khurshid Ali");

Console.WriteLine(JsonSerializer.Serialize(jsonDoc));

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionPaul GrimshawView Question on Stackoverflow
Solution 1 - C#ClintView Answer on Stackoverflow
Solution 2 - C#Paul GrimshawView Answer on Stackoverflow
Solution 3 - C#Khurshid AliView Answer on Stackoverflow