Passing complex navigation parameters with MvvmCross ShowViewModel
Asked Answered
F

2

15

My complex type wouldn't pass from Show to Init method even with configured MvxJsonNavigationSerializer as specified here Custom types in Navigation parameters in v3

public class A
{
 public string String1 {get;set;}
 public string String2 {get;set;}
 public B ComplexObject1 {get;set;}
}

public class B
{
 public double Double1 {get;set;}
 public double Double2 {get;set;}
}

When I pass instance of object A to ShowViewModel method I receive this object with String1 & String2 deserialized correctly but CopmlexObject1 is null.

How to deal with complex object MvvmCross serialization?

Forty answered 27/9, 2013 at 18:59 Comment(1)
I was able to resolve this issue by adding mvvmcross json plugin in my uiview project.Raymundorayna
Z
26

I believe there may be some gremlins in that previous answer - will log as an issue :/


There are other possible routes to achieve this type of complex serializable object navigation still using Json and overriding parts of the framework, but actually I think that it might be better to just use your own BaseViewModel's to do serialization and deserialization - e.g. use serialization code like:

public class BaseViewModel
    : MvxViewModel
{
    private const string ParameterName = "parameter";

    protected void ShowViewModel<TViewModel>(object parameter)
        where TViewModel : IMvxViewModel
    {
        var text = Mvx.Resolve<IMvxJsonConverter>().SerializeObject(parameter);
        base.ShowViewModel<TViewModel>(new Dictionary<string, string>()
            {
                {ParameterName, text}
            });
    }
}

with deserialization like:

public abstract class BaseViewModel<TInit>
    : MvxViewModel
{
    public void Init(string parameter)
    {
        var deserialized = Mvx.Resolve<IMvxJsonConverter>().DeserializeObject<TInit>(parameter);
        RealInit(deserialized);
    }

    protected abstract void RealInit(TInit parameter);
}

then a viewModel like this:

public class FirstViewModel
    : BaseViewModel
{
    public IMvxCommand Go
    {
        get
        {
            return new MvxCommand(() =>
                {
                    var parameter = new A()
                        {
                            String1 = "Hello",
                            String2 = "World",
                            ComplexObject = new B()
                                {
                                    Double1 = 42.0,
                                    Double2 = -1
                                }
                        };
                    ShowViewModel<SecondViewModel>(parameter);
                });
        }
    }
}

can navigate to something like:

public class SecondViewModel
    : BaseViewModel<A>
{
    public A A { get; set; }

    protected override void RealInit(A parameter)
    {
        A = parameter;
    }
}
Zahavi answered 27/9, 2013 at 20:56 Comment(1)
Thank you! Interesting approach and that will definitely work because I'm fully control serialization/deserialization process by myself.Forty
A
10

A small addition to Stuart's answer to add type safety:

public class BaseViewModel: MvxViewModel {

    protected bool ShowViewModel<TViewModel, TInit>(TInit parameter) where TViewModel: BaseViewModel<TInit> {
        var text = Mvx.Resolve<IMvxJsonConverter>().SerializeObject(parameter);
        return base.ShowViewModel<TViewModel>(new Dictionary<string, string> { {"parameter", text} });
    }
}

public abstract class BaseViewModel<TInit> : BaseViewModel {

    public void Init(string parameter)
    {
        var deserialized = Mvx.Resolve<IMvxJsonConverter>().DeserializeObject<TInit>(parameter);
        RealInit(deserialized);
    }

    protected abstract void RealInit(TInit parameter);
}

ShowViewModel method now takes the same parameter type that the RealInit method instead of an object type. Also, BaseViewModel<TInit> inherits from BaseViewModel so their instances can also call the new ShowViewModel method.

The only drawback is that you have to explicitly specify the parameter type in the call like this:

ShowViewModel<StoreInfoViewModel, Store>(store);
Adena answered 6/4, 2015 at 13:10 Comment(0)

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