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Arjan
365
Bind ContextMenu to columns of grid
posted

Hello,

I want to bind the columns of a grid to a contextmenu so I can change te visibility of the columns by a checkbox. On right click i want the ContextMenu to open and show a list of menuitems with checkboxes for each column. I want to do this in XAML.

I tried the following with no luck:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

<

 

 

igMenu:ContextMenuService.Manager

>

 

 

 

<!--Set the attached Manager property to an instance of a ContextMenuManager object-->

 

 

 

<igMenu:ContextMenuManager OpenMode

="RightClick">

 

 

 

<!--Set the ContextMenuManager object's ContextMenu property to an instance of a xamWebContextMenu control-->

 

 

 

<igMenu:ContextMenuManager.ContextMenu

>

 

 

 

<igMenu:XamWebContextMenu ItemsSource="{Binding Source=xRelationGrid, Path

=Columns}">

 

 

 

<igMenu:XamWebContextMenu.ItemTemplate

>

 

 

 

<DataTemplate

>

 

 

 

<Grid

>

 

 

 

<CheckBox IsChecked="{Binding Path=Visibility,

 

 

Converter={StaticResource VisibilityToBool}}" Content="{Binding Path=Key}" />

 

 

 

</Grid>

 

 

 

</DataTemplate

>

 

 

 

</igMenu:XamWebContextMenu.ItemTemplate

>

 

 

 

</igMenu:XamWebContextMenu

>

 

 

 

</igMenu:ContextMenuManager.ContextMenu

>

 

 

 

</igMenu:ContextMenuManager

>

 

 

 

</igMenu:ContextMenuService.Manager

>

Thanx,

Arjan

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  • 3071
    posted

    Hi Arjan,
    You can do this in XAML but you need to inherit from XamWebContextMenu:
    public class GridContextMenu : XamWebContextMenu
    {
      public static readonly DependencyProperty TheGridProperty =
        DependencyProperty    .Register("TheGrid", typeof(XamWebGrid),
            typeof(GridContextMenu), new PropertyMetadata(null));
      public XamWebGrid TheGrid
      {
        get{return (XamWebGrid)this.GetValue(GridContextMenu.TheGridProperty);}
        set{this.SetValue(GridContextMenu.TheGridProperty, value);}
      }
      protected override void OnOpening(OpeningEventArgs args)
      {
        this.TheGrid = this.PlacementTargetResolved as XamWebGrid;
        base.OnOpening(args);
      }
    }
    This sample shows how to use the GridContextMenu in XAML:
    <igGrid:XamWebGrid x:Name="TestGrid">
      <igMenu:ContextMenuService.Manager>
        <igMenu:ContextMenuManager>
          <igMenu:ContextMenuManager.ContextMenu>
            <l:GridContextMenu 
              ItemsSource="{Binding RelativeSource={RelativeSource Self}, Path=TheGrid.Columns}">
              <l:GridContextMenu.DefaultItemsContainer>
                <DataTemplate>
                  <igMenu:XamWebMenuItem Header="{Binding}" 
                        IsCheckable    ="True"
                            IsChecked="{Binding Path=Visibility, Converter={StaticResource VisibilityConverter}, Mode=TwoWay}">
                  </igMenu:XamWebMenuItem>
                </DataTemplate>
              </l:GridContextMenu.DefaultItemsContainer>
            </l:GridContextMenu>
          </igMenu:ContextMenuManager.ContextMenu>
        </igMenu:ContextMenuManager>
      </igMenu:ContextMenuService.Manager>
    </    igGrid:XamWebGrid>
    Regards,
    Marin 

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