Question

在Silverlight和PRISM中，通过从不同模块中的ViewModel传递参数，在一个模块中打开弹出子窗口的好方法是什么。

Answer 1

创建一个名为IChildWindowService的模块已知的公共接口/类，在引导程序中注册IChildWindowServe / ChildWindowService。

//Highly simplified version
//Can be improved by window reuse, parameter options, stronger eventing
public class ChildWindowService : IChildWindowService
{
    public ChildWindowService(IServiceLocator container)
    {
        _container = container;
    } 
    public void Show<TViewModel>(TViewModel viewModel = null, Action<TViewModel, bool?> callBack = null) where TViewModel is IViewModel
    {
        var viewName = typeof(TViewModel).Name.Replace("Model", string.Empty);
        // In bootstrapper register all instances of IView or register each view one by one
        var view = _container.GetInstance<IView>(viewName);
        viewModel = viewModel ?? _container.GetInstance<TViewModel>();
        view.DataContext = viewModel;
        var window = new ChildWindow();
        window.Content = view;
        var handler = (s,e) => { window.Close(); }
        viewModel.RequestClose += handler; 
        view.Closed += (s,e) => { viewModel.RequestClose -= handler; }
        // In silverlight all windows show as Modal, if you are using a third party you can make a decision here
        window.Show();
    }
}

创建一个通用的CompositePresentationEvent，此事件将参数从a点传递到b点

public class OpenChildWindowWithParameters : CompositePresentationEvent<ParamEventArgs>{}

模块A中的ViewModel引发了事件。模块B中的控制器注册并响应事件。模块B中的Controller将子窗口服务作为依赖项。当事件被引发时，Controller将在模块B中创建VM并将参数传递给它，从事件中，它还将使用服务来显示ChildWindow。

Silverlight 4和Prism中的弹出窗口

1 个答案: