I am new to MVVM .I have a window Demo.xaml which has menu, under menu i have sub menus.I want to open sub menu window i.e Test.Xaml on the click of submenu using MVVM approach. I made an object of Test Window but it didn't showed "Show" property. I also tried by using Delegate Commands but i failed. How can I achieve this?
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When you say MVVM approach, do you mean the ViewModel should be able to spawn a new window?grek40– grek402018-01-23 08:10:32 +00:00Commented Jan 23, 2018 at 8:10
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2 Answers
The most common MVVM approach to solve your problem is to bind the submenu item's "Command" to a property in the ViewModel which implements ICommand. Then, in the command's execution, you can open your window of choice.
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You need to use ICommand - when working with MVVM using RelayCommand is best practice.
See my example below:
MainWindow.xaml
<MenuItem Header="Settings" Command="{Binding CmdOpenSetting}" >
<MenuItem.Icon>
<Image Source="..\Resources\if_Gnome-Preferences-System.png" Height="16" Width="16" Margin="0,0,-5,0" />
</MenuItem.Icon>
</MenuItem>
MainWindow.xaml.cs
public MainWindow()
{
InitializeComponent();
//Singelton not needed
MainWindowViewModel.Instance = new MainWindowViewModel();
this.DataContext = MainWindowViewModel.Instance;
//Also works
this.DataContext = new MainWindowViewModel();
}
MainWindowViewModel.cs
private RelayCommand _commandOpenSettings;
public ICommand CmdOpenSetting
{
get
{
if(_commandOpenSettings.IsNull())
{
_commandOpenSettings = new RelayCommand(param => OpenSettings());
}
return _commandOpenSettings;
}
}
Note you may have to adjust the CmdOpenSetting according to you implementation of RelayCommand
