从WPF / MVVM应用程序的主窗口中打开第二个窗口

Question

Could you tell me how to in pure MVVM way call (I mean open/show) child window from parent window. Let's say I have two Views:

1. MainWindow.cs (MainWindow.xaml) - parent window (DataContext = new MainWindowViewModel())
2. Window.cs (Window.xaml) - child window (DataContext = new WindowViewModel())

And corresponding ViewModel classes:

1. MainWindowViewModel.cs
2. WindowViewModel.cs

I would like my window to be opened after button click (button that is on the MainWindow view). Because of that I have defined command binding in MainWindow.xaml:

<Button x:Name="buttonOpenWindow" Content="Open window..." Width="100" Height="20" Command="{Binding OpenWindowCmd}"/>

And MainWindowViewModel.cs piece:

public ICommand OpenWindowCmd { get; set; }

public MainWindowViewModel()
{
    OpenWindowCmd = new RelayCommand(o => OpenWindow());
}

private void OpenWindow()
{
    // What to put here?
}

In Window.xaml I added something like that:

<Window x:Class="Namespace.View.Window"
    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"
    mc:Ignorable="d"
    xmlns:vm="clr-namespace:Namespace.ViewModel"
    Title="Title" Height="300" Width="325" Visibility="{Binding IsWindowVisible, Converter={StaticResource BooleanToVisibilityConverter}}">
<Window.Resources>
    <BooleanToVisibilityConverter x:Key="BooleanToVisibilityConverter"/>
</Window.Resources>

(...)

And the WindowViewModel.cs:

using System.ComponentModel;
using System.Runtime.CompilerServices;
using Namespace.Annotations;

namespace Namespace.ViewModel
{
    public class WindowViewModel : INotifyPropertyChanged
    {
        private bool _isWindowVisible;
        public bool IsWindowVisible
        {
            get { return _isWindowVisible; }
            set
            {
                _isWindowVisible = value;
                OnPropertyChanged(nameof(IsWindowVisible));
            }
        }

        public event PropertyChangedEventHandler PropertyChanged;

        [NotifyPropertyChangedInvocator]
        protected virtual void OnPropertyChanged([CallerMemberName] string propertyName = null)
        {
            PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(propertyName));
        }
    }
}

I am not sure what to do next and if that approach is correct. I found some services implementations in the forum, but I thought of using just Visibility property instead (but not sure if it is possible). I need to somehow change the IsWindowVisible in one of the view models I suppose. Could anyone suggest how to gently handle such sub window opening?

Answer 1

If I understood well, you need something like this:

private void OpenWindow()
{
    WindowViewModel wvm = new WindowViewModel();
    Window win = new Window()
    {
        DataContext = wvm;
    };
    win.Show();
}

If you don't like this solution then try the one from the comments with IWindowService. In any case it makes no sense to use a Visibility property.