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我被要求将使用 MVVM 模式的独立 WPF 应用程序转换为用户控件。这个应用程序由一个主窗口和几个其他窗口组成。但是，我在尝试这样做时遇到了一些错误，这些错误都指向我的 App.xaml 类和声明为转换器的其他资源：

Library project file cannot specify ApplicationDefintion element. The project file contains a property value that is not valid. The name "ViewModelLocator" does not exist in the namespace "clr-namespace:MapperX.ViewModels".

所以看起来错误围绕着我的 ViewModelLocator。

目前项目目录结构设置如下：

顶级 -> ViewModels 文件夹 -> ViewModelLocator

App.xaml 的设置如下：

<Application x:Class="MapperX.App"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:local="clr-namespace:MapperX"
             xmlns:vm="clr-namespace:MapperX.ViewModels"
             StartupUri="MainWindow.xaml">
    <Application.Resources>
        <vm:ViewModelLocator x:Key="ViewModelLocator" />
    </Application.Resources>
</Application>

这是该ViewModelLocator课程的一个片段：

namespace MapperX.ViewModels
{
    /// <summary>
    /// This class instantiates all the viewmodels
    /// </summary>
    public class ViewModelLocator
    {
        WpfMap map = new WpfMap();

        private MainViewModel _mainViewModel;
        public MainViewModel MainViewModel
        {
            get
            {
                if (_mainViewModel == null)
                {
                    _mainViewModel = new MainViewModel(map)
                }
                return _mainViewModel;
            }
        }

        private LayersViewModel _layersViewModel;
        public LayersViewModel LayersViewModel
        {
            get
            {
                if (_layersViewModel == null)
                {
                    _layersViewModel = new LayersViewModel(map)
                }
                return _layersViewModel;
            }
        }
    }
}

然后我DataContext为视图 .xaml 设置如下： DataContext="{Binding Path=MainViewModel, Source={StaticResource ViewModelLocator}}"

在没有 App.xaml 的情况下仍然能够使用 ViewModelLocator 的正确方法是什么？

4

2 回答 2

1

DataContext您可以创建附加属性来设置控件库中的视图，而不是依赖资源：

namespace ControlsAndResources
{
    public class View
    {
        private static readonly ViewModelLocator s_viewModelLocator = new ViewModelLocator();

        public static readonly DependencyProperty ViewModelProperty = DependencyProperty.RegisterAttached("ViewModel", typeof(string), 
            typeof(ViewModelLocator), new PropertyMetadata(new PropertyChangedCallback(OnChanged)));

        public static void SetViewModel(UserControl view, string value) => view.SetValue(ViewModelProperty, value);

        public static string GetViewModel(UserControl view) => (string)view.GetValue(ViewModelProperty);

        private static void OnChanged(DependencyObject d, DependencyPropertyChangedEventArgs e)
        {
            UserControl view = (UserControl)d;
            string viewModel = e.NewValue as string;
            switch (viewModel)
            {
                case "MainViewModel":
                    view.DataContext = s_viewModelLocator.MainViewModel;
                    break;
                case "LayersViewModel":
                    view.DataContext = s_viewModelLocator.LayersViewModel;
                    break;
                default:
                    view.DataContext = null;
                    break;
            }
        }
    }
}

用法：

<UserControl xmlns:local="clr-namespace:ControlsAndResources" ... 
             local:View.ViewModel="MainViewModel">
于 2019-01-10T14:15:49.813 回答
1

为什么不把它ViewModelLocator放在你的顶层UserControl？

<UserControl.Resources>
   <vm:ViewModelLocator x:Key="ViewModelLocator" />
</UserControl.Resources>
于 2019-01-10T13:32:47.870 回答