将项目添加到 ObservableCollection 并使其在其绑定的 ListCollectionView 中被选中

Question

我有一个带有 ObservableCollection 的 ViewModel，以及一个使用 Xaml 声明的 CollectionViewSource 的视图，它绑定到 DataGrid.ItemsSource.

ViewModel 有一个创建新对象并将其添加到 ObservableCollection 的命令。同样在 ViewModel 中，我正在处理 CollectionChanged 事件，因此当发生添加时我 "know"。

我的疑问是：

how can I make the just-added object selected in the DataGrid?

请注意，添加和更改事件都在 ViewModel 中，但选择必须在 View 中发生。

我不介意基于代码隐藏的解决方案。

视图模型：

public class TelaInicialViewModel : ViewModelBase
{
    public ObservableCollection<PacienteViewModel> Pacientes { get; set; }

    IRepositório<Paciente> _repositório_pacientes = new RepositórioPaciente();

    // CONSTRUTOR
    public TelaInicialViewModel()
    {
        var pacientes = _repositório_pacientes.Items.Select(p => new PacienteViewModel(p));
        Pacientes = new ObservableCollection<PacienteViewModel>(pacientes);
        Pacientes.CollectionChanged += Pacientes_CollectionChanged;
    }


    void Pacientes_CollectionChanged(object sender, NotifyCollectionChangedEventArgs e)
    {
        switch (e.Action)
        {
            case NotifyCollectionChangedAction.Add:
                foreach (var added in e.NewItems)
                {
                    var p = added as PacienteViewModel;
                    if (p != null)
                    {
                        var novoPaciente = p.GetModel();  /////////  HERE I HAVE IT!!
                        _repositório_pacientes.Adicionar(novoPaciente);
                    }
                }
                break;
            case NotifyCollectionChangedAction.Replace:
                // ...
                break;
        }
    }


    void AdicionarPaciente()
    {
        var formulárioPaciente = new FormulárioCriaçãoPaciente();
        PacienteViewModel novoPaciente = formulárioPaciente.ExibirDiálogo(new Paciente());
        if (novoPaciente != null)
            //////// HERE I HAVE IT!!!
            Pacientes.Add(novoPaciente);
    }
    public ICommand ComandoAdicionarPaciente { get { return new RelayCommand(AdicionarPaciente); } }
}

查看 (Xaml)

<UserControl.Resources>

    <CollectionViewSource
        x:Key="PacientesViewSource"
        Source="{Binding Pacientes}"
        Filter="PacientesViewSource_Filter"/>

</UserControl.Resources>


<DockPanel>

    <DockPanel DockPanel.Dock="Top" Margin="0,10,0,0" >
        <TextBox x:Name="FiltroPacienteTextBox" Height="30"
            TextChanged="FiltroPacienteTextBox_TextChanged"
            DockPanel.Dock="Bottom" Margin="10,10,10,0"/>
        <Button x:Name="botaoNovoPaciente" Width="120" Content="Novo"
            HorizontalAlignment="Left" Height="30" Margin="10,0,0,0"
            Command="{Binding ComandoAdicionarPaciente}"/>
    </DockPanel>

    <DataGrid x:Name="pacienteDataGrid" Margin="0,10,0,0"
        AutoGenerateColumns="False"
        CanUserAddRows="False" CanUserDeleteRows="False"
        CanUserReorderColumns="False" CanUserResizeRows="False"
        ItemsSource="{Binding Source={StaticResource PacientesViewSource}}"
        IsSynchronizedWithCurrentItem="True">
    </DataGrid>

</DockPanel>

Answer 1

嗯，看了一些评论建议的答案，我发现了一个很方便的方法来做我想做的事：

查看（代码隐藏）

public partial class TelaInicialView : UserControl
{
    CollectionViewSource _pacientes_source;

    public TelaInicialView()
    {
        InitializeComponent();

        Loaded += TelaInicialView_Loaded;
    }


    void TelaInicialView_Loaded(object sender, RoutedEventArgs e)
    {
        _pacientes_source = FindResource("PacientesViewSource") as CollectionViewSource;

        // FIRST STEP: CAST CollectionView.Source to INotifyCollectionChanged, and listen to its CollectionChanged event
        var source = _pacientes_source.Source as INotifyCollectionChanged;
        source.CollectionChanged += TelaInicialView_CollectionChanged;
    }

    void TelaInicialView_CollectionChanged(object sender, NotifyCollectionChangedEventArgs e)
    {
        // SECOND STEP: WHEN YOU ADD SOMETHING, SELECT IT!
        if (e.Action == NotifyCollectionChangedAction.Add)
            pacienteDataGrid.SelectedItem = e.NewItems[0];
    }

    // ...
}

Answer 2

我鼓励您在这里坚持使用 MVVM 模式。在这种情况下，我通常会向 ViewModel 添加表示所选项目的属性。这会将所有重要逻辑保留在 ViewModel 中，以便对其进行测试。我会在您的 ViewModel 中添加如下内容（如果您的 ViewModel 是 DependencyObject，您可以在此处使用 DependencyProperty）：

private Paciente _SelectedPaciente;
public Paciente SelectedPaciente
{
    get { return this._SelectedPaciente; }
    set
    {
        if (this._SelectedPaciente != value)
        {
            this._SelectedPaciente = value;
            this.RaisePropertyChanged("SelectedPaciente");
        }
    }
}

然后像这样修改你的DataGrid：

<DataGrid x:Name="pacienteDataGrid" 
          Margin="0,10,0,0"
          AutoGenerateColumns="False"
          CanUserAddRows="False"
          CanUserDeleteRows="False"
          CanUserReorderColumns="False"
          CanUserResizeRows="False"
          ItemsSource="{Binding Source={StaticResource PacientesViewSource}}"
          SelectedItem="{Binding SelectedPaciente, Mode=TwoWay}"
          IsSynchronizedWithCurrentItem="True">

然后像这样修改你的AdicionarPaciente：

void AdicionarPaciente()
{
    var formulárioPaciente = new FormulárioCriaçãoPaciente();
    PacienteViewModel novoPaciente = formulárioPaciente.ExibirDiálogo(new Paciente());
    if (novoPaciente != null)
    {
        //////// HERE I HAVE IT!!!
        Pacientes.Add(novoPaciente);
        this.SelectedPaciente = novoPaciente;
    }
}

现在您可以为您的 ViewModel 编写单元测试以确认当调用添加新 Paciente 的命令时新的 Paciente 被选中。

将项目添加到 ObservableCollection 并使其在其绑定的 ListCollectionView 中被选中

Adding item to ObservableCollection and make it selected in its bound ListCollectionView

wpf

mvvm

observablecollection

selecteditem

collectionview

视图模型：

查看 (Xaml)

查看（代码隐藏）