The parameter conversion from type 'System.String' to type ''X' failed because no type converter can convert between these types

Question

I\'m stumped with this one and your help would be most appreicated.

I get the error:

The parameter conversion from type \'System.String\' to t

Answer 1

I had the same error but the cause was different from the accepted answer. Adding a custom ModelBinder fixed it for me.

I created a CData class as described here

How do you serialize a string as CDATA using XmlSerializer? (it's the last answer unless it got voted up because it's a great solution)

I serialize my ViewModel to XML. When I set a type in a property in my ViewModel to CData, it will automatically be serialized to a CData section and deserialized to a CData field. (which is great!)

When I do a post action in my view I want to keep some fields in my ViewModel so they are added as Html.HiddenFor in the form.

When a submit is done, an error occurs (the one in the title). The model binder tries to bind a string field to a CData field which fails.

A custom model binder for the CData type fixes this

public class CDataModelBinder : IModelBinder
{
    public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext)
    {
        var result = bindingContext.ValueProvider.GetValue(bindingContext.ModelName);
        return new CData(result.AttemptedValue);
    }
}

Global.asax

 ModelBinders.Binders.Add(typeof(CData), new CDataModelBinder());

Answer 2

The problem is the name of your action parameter:

public virtual ActionResult Create(FeedbackComment comment)

It's called comment. But your FeedbackComment also has a property called Comment of type string. So the default model binder gets crazy. Just rename one of the two to avoid the conflict.

For example the following will fix your issue:

public virtual ActionResult Create(FeedbackComment model)
{
    if (ModelState.IsValid)
    {
        _feedbackCommentRepository.Add(model);
        return RedirectToAction(MVC.Feedback.Details(model.FeedbackId));
    }
    return View(model);
}

Answer 3

I had the same name for the parameter on the GET method

[HttpGet]
public ActionResult Create(int OSSB)

.. and a property from Faturamento model, that was used on the POST method

[HttpPost]
public ActionResult Create(Faturamento model)

Just like this..

public class Faturamento {
        [Column("ID_OSSB")]
        public virtual int ID_OSSB { get; set; }
        [ForeignKey("ID_OSSB")]
        public virtual OSSB OSSB { get; set; }
        ...
}

What solved for me was changing the GET parameter name:

[HttpGet]
public ActionResult Create(int IDOSSB)

Answer 4

In my case I was writing a model class name in Html Helper class. which was getting it failed as it was expecting a System.String datatype and I was passing a custom class name.