Displaying Popup in WPF using MVVM

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Hi I am using MVVM first time, and worried about displaying Popup control of my XAML using ViewModel, My XMAL is as follows-

 <Button Content="Add Customer"  Name = "AddButton" Command="{Binding ShowPopup, UpdateSourceTrigger=PropertyChanged }"/>

  Popup Name="TestPopUp" IsOpen="False"
   StackPanel Margin="0,0,0,30"
    TextBlock Text="First Name:" FontWeight="Bold" Margin="0,0,5,0"
     TextBlock Text="{Binding FirstName}" MinWidth="50" Margin="0,0,5,0"
     TextBlock Text="Last Name:" FontWeight="Bold" Margin="0,0,5,0"
     TextBlock Text="{Binding LastName}" MinWidth="50" Margin="0,0,5,0"
     TextBlock Text="Age:" FontWeight="Bold" Margin="0,0,5,0"
     TextBlock Text="{Binding Age}" MinWidth ="150"  Margin="0,0,5,0"
     Button Content="Add Customer" Command="{Binding AddCustomerCommand}"Height="27" MinWidth ="200" Width="200"
                StackPanel
            Popup

My view Model Code:-
 public class MainViewModel : ViewModelBase
    {

        private DelegateCommand<object> _showpopup;

        private void DisplayPopup(object param)
        {

            MessageBox.Show("Pop up need to be open Here");

        }


   public DelegateCommand<object> ShowPopup
        {
            get
            {
                if (_showpopup  == null)
                {
                    _showpopup = new DelegateCommand<object>(DisplayPopup, CanAddCustomer);
                }
                return _showpopup;
            }
        }

the alert Massage is showing correctly , but I wat to open the Popup window instead. Please help.

Posted 14-Jul-11 21:37pm

vinod deo from navi mumbai

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Pete O'Hanlon · Accepted Answer · 2011-07-14T22:05:00

The standard way to achieve this in MVVM is actually very straightforward. All you need to do is add a boolean property to your ViewModel that you bind the IsOpen property to on your Popup. When you raise the change notification, the Popup should react appropriately (as long as you haven't changed the default update trigger). Here's what you need to add to your ViewModel (assuming that your VM uses RaisePropertyChanged as the method name for property notification):

C#

private bool _isOpen;
public bool IsOpen
{
  get { return _isOpen; }
  set
  {
    if (_isOpen == value) return;
    _isOpen = value;
    RaisePropertyChanged("IsOpen");
  }
}

In your Popup declaration in the XAML, add the following attribute

XML

IsOpen="{Binding IsOpen}"

And that's it - an MVVM friendly way to show the popup.