DataControlBase.ColumnGeneratorTemplate Property

Gets or sets a template that describes grid columns. This is a dependency property.

Namespace: DevExpress.Xpf.Grid

Assembly: DevExpress.Xpf.Grid.v23.2.Core.dll

NuGet Package: DevExpress.Wpf.Grid.Core

Declaration

C#

public DataTemplate ColumnGeneratorTemplate { get; set; }

VB.NET

Public Property ColumnGeneratorTemplate As DataTemplate

Property Value

Type	Description
DataTemplate	The column template.

Remarks

You can define columns in a ViewModel and display them in the GridControl. To do this, follow the steps below:

1. Create a collection of grid columns in a ViewModel.

2. Specify a data template that generates columns:

   XAML

   <Window.Resources>
       <DataTemplate x:Key="DefaultColumnTemplate">
           <dxg:GridColumn FieldName="{Binding Path=(dxci:DependencyObjectExtensions.DataContext).FieldName, RelativeSource={RelativeSource Self}}"/>
       </DataTemplate>
   </Window.Resources>
   

3. Assign the column collection to the ColumnsSource property and the column template to the ColumnGeneratorTemplate property:

   XAML

   <dxg:GridControl ...
                    ColumnsSource="{Binding Columns}" 
                    ColumnGeneratorTemplate="{StaticResource DefaultColumnTemplate}"/>
   

If you have multiple column templates, use the ColumnGeneratorTemplateSelector property to specify a template selector that chooses a template based on a column type.

Refer to the following help topic for more information: How to: Bind the Grid to a Collection of Columns.