Question

我已经使用Windows Template Studio创建了MVVM Light应用程序，现在我正尝试在XAML中使用数据绑定。但这对生成的代码来说似乎是不可能的，因为它没有在XAML中声明Viewmodel。我更改了代码以执行此操作，但是随后我对这个错误感到沮丧：

不能构造XAML ViewModelLocator类型。为了在XAML中进行构造，类型不能为抽象，接口，嵌套，泛型或结构，并且必须具有公共默认构造函数

我尝试将构造函数公开，但这不能解决问题。关于如何使数据绑定工作的任何想法？

这是所有代码：

App.XAML

<Application
x:Class="PatientApp.UWP.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:vm="using:PatientApp.UWP.ViewModels">

<Application.Resources>
    <ResourceDictionary>
        <ResourceDictionary.MergedDictionaries>
            <XamlControlsResources  xmlns="using:Microsoft.UI.Xaml.Controls"/>
            <ResourceDictionary Source="/Styles/_Colors.xaml"/>
            <ResourceDictionary Source="/Styles/_FontSizes.xaml"/>
            <ResourceDictionary Source="/Styles/_Thickness.xaml"/>
            <ResourceDictionary Source="/Styles/TextBlock.xaml"/>
            <ResourceDictionary Source="/Styles/Page.xaml"/>
        </ResourceDictionary.MergedDictionaries>
        <vm:ViewModelLocator xmlns:vm="using:PatientApp.UWP.ViewModels" x:Key="Locator" />
    </ResourceDictionary>
</Application.Resources>

ViewModelLocator：

    [Windows.UI.Xaml.Data.Bindable]
public class ViewModelLocator
{
    private static ViewModelLocator _current;

    public static ViewModelLocator Current => _current ?? (_current = new ViewModelLocator());

    private ViewModelLocator()
    {
        SimpleIoc.Default.Register(() => new NavigationServiceEx());
        SimpleIoc.Default.Register<ShellViewModel>();
        Register<PatientsViewModel, AllPatientsPage>();
        Register<DiseasesViewModel, AllDiseasesPage>();
    }

    public DiseasesViewModel DiseasesViewModel => SimpleIoc.Default.GetInstance<DiseasesViewModel>();

    public PatientsViewModel PatientsViewModel => SimpleIoc.Default.GetInstance<PatientsViewModel>();

    public ShellViewModel ShellViewModel => SimpleIoc.Default.GetInstance<ShellViewModel>();

    public NavigationServiceEx NavigationService => SimpleIoc.Default.GetInstance<NavigationServiceEx>();

    public void Register<VM, V>()
        where VM : class
    {
        SimpleIoc.Default.Register<VM>();

        NavigationService.Configure(typeof(VM).FullName, typeof(V));
    }
}

PatientPage：

<Page
x:Class="PatientApp.UWP.Views.AllPatientsPage"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
Style="{StaticResource PageStyle}"
mc:Ignorable="d"
xmlns:datamodel="using:PatientApp.Model"
DataContext="{Binding PatientViewModelInstance, Source={StaticResource Locator}}">
<Grid
        Background="{ThemeResource SystemControlPageBackgroundChromeLowBrush}">
    <Grid.RowDefinitions>
        <RowDefinition Height="Auto" />
        <RowDefinition Height="*" />
    </Grid.RowDefinitions>

    <CommandBar DefaultLabelPosition="Right"
                Grid.Row="0"
                OverflowButtonVisibility="Collapsed">

        <AppBarButton Icon="Sort" Label="Sort" />
        <AppBarButton Icon="Edit" Label="Edit" />
        <AppBarButton Icon="Add" Label="Add" />
        <AppBarButton Icon="Zoom" Label="Search" />
    </CommandBar>
    <GridView  x:Name="Patients"
               Grid.Row="1"
               ItemsSource="{Binding Patients}">
        <GridView.ItemsPanel>
            <ItemsPanelTemplate>
                <ItemsStackPanel Margin="14,0,0,0" Orientation="Vertical" />
            </ItemsPanelTemplate>
        </GridView.ItemsPanel>
        <GridView.ItemTemplate>
            <DataTemplate x:DataType="datamodel:Patient">
                <TextBlock Text="{x:Bind Name}"
                                   FontWeight="Medium"
                                   TextWrapping="NoWrap"
                                   HorizontalAlignment="Left" />
            </DataTemplate>
        </GridView.ItemTemplate>
    </GridView>
</Grid>

Answer 1

不能构造XAML ViewModelLocator类型。为了在XAML中进行构造，类型不能为抽象，接口，嵌套，泛型或结构，并且必须具有公共默认构造函数

原因是ViewModelLocator构造方法在您的方案中是私有的。因此，您无法在xaml中实例化它。您可以将其更改为公开。即使您不能使用它。因为xaml中的ViewModelLocator实例是由构造函数而不是单例方法创建的。换句话说<vm:ViewModelLocator xmlns:vm="using:PatientApp.UWP.ViewModels" x:Key="Locator" />不等于ViewModelLocator.Current。不幸的是，ActivationService使用ViewModelLocator.Current来获得NavigationService。因此，应用在启动时会引发异常。

public static NavigationServiceEx NavigationService => ViewModelLocator.Current.NavigationService;

使用MVVMLight很难解决。通常，页面DataContext被后面的代码保护。并使用x：bind绑定到绑定数据源。

private SettingsViewModel ViewModel
{
    get { return ViewModelLocator.Current.SettingsViewModel; }
}

WTS无法在XAML中声明ViewModelLocator

1 个答案: