In the neighbour post: How should the ViewModel close the form? I've posted my vision how to close windows with MVVM usage. And now I have a question: how to open them.
I have a main window (main view). If user clicks on the "Show" button then "Demo" window (modal dialog) should be displayed. What is a preferable way to create and open windows using MVVM pattern? I see two general approaches:
The 1st one (probably the simplest). Event handler "ShowButton_Click" should be implemented in the code behind of the main window in way like this:
private void ModifyButton_Click(object sender, RoutedEventArgs e)
{
ShowWindow wnd = new ShowWindow(anyKindOfData);
bool? res = wnd.ShowDialog();
if (res != null && res.Value)
{
// ... store changes if neecssary
}
}
- If we "Show" button state should be changed (enabled/disabled) we will need to add logic that will manage button state;
- The source code is very similar to "old-style" WinForms and MFC sources - I not sure if this is good or bad, please advise.
- Something else that I've missed?
Another approach:
In the MainWindowViewModel we will implement "ShowCommand" property that will return ICommand interface of the command. Comman in turn:
- will raise "ShowDialogEvent";
- will manage button state.
This approach will be more suitable for the MVVM but will require additional coding: ViewModel class can't "show dialog" so MainWindowViewModel will only raise "ShowDialogEvent", the MainWindowView we will need to add event handler in its MainWindow_Loaded method, something like this:
((MainWindowViewModel)DataContext).ShowDialogEvent += ShowDialog;
(ShowDialog - similar to the 'ModifyButton_Click' method.)
So my questions are: 1. Do you see any other approach? 2. Do you think one of the listed is good or bad? (why?)
Any other thoughts are welcome.
Thanks.