No data with ListBox with Grid inside - xaml

Question

I am doing a Grid of two columns that are inside a ListBox. That after that I can DataBind the two columns to be repeated vertically.
So far the code below shows nothing on the WP7 emulator.
<ListBox Background="Yellow" ItemsSource="{Binding}" Height="100" Margin="0,0,8,0">
<ListBox.ItemTemplate>
<DataTemplate>
<Grid Height="100">
<Grid.ColumnDefinitions>
<ColumnDefinition Width="200" />
<ColumnDefinition Width="200" />
</Grid.ColumnDefinitions>
<TextBlock Text="Channels" HorizontalAlignment="Stretch" Foreground="Black" Grid.Column="0" />
<TextBlock Text="Antenna" HorizontalAlignment="Stretch" Foreground="Black" Grid.Column="1"/>
</Grid>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
Please help me.

Answer 1

If your only concern is that you see ItemTemplate in action, you can supply explicit non-UI items as follows:
<ListBox Background="Yellow" Height="100" Margin="0,0,8,0" xmlns:sys="clr-namespace:System;assembly=mscorlib">
<ListBox.ItemTemplate>
<DataTemplate>
<Grid Height="30">
<Grid.ColumnDefinitions>
<ColumnDefinition Width="200" />
<ColumnDefinition Width="200" />
</Grid.ColumnDefinitions>
<TextBlock Text="Channels" HorizontalAlignment="Stretch" Foreground="Black" Grid.Column="0" />
<TextBlock Text="Antenna" HorizontalAlignment="Stretch" Foreground="Black" Grid.Column="1"/>
</Grid>
</DataTemplate>
</ListBox.ItemTemplate>
<sys:String>1111111</sys:String>
<sys:String>2222222</sys:String>
<sys:String>3333333</sys:String>
</ListBox>
Notes:
I removed ItemsSource and supplied items explicitly.
Items must not derive from UIElement so that they are templated. (UIElements are simply drawn and the template is ignored.)
I added System namespace so that string objects can be specified.
I decreased ItemTemplate height so that more than one list row is visible.
Easier solution:
Give the ListBox a name and remove the binding:
<ListBox x:Name="myLB" Background="Yellow" Height="100" Margin="0,0,8,0">
Then use this line in the code (after the call InitializeComponent()):
myLB.ItemsSource = new List<string> { "First", "Second", "Third" };

Answer 2

If you want design-time itemssource, you can use the IsInDesignMode property like so:
if (System.ComponentModel.DesignerProperties.IsInDesignTool)
{
myListBox.ItemsSource = GenerateMockItems();
}
else
{
myListBox.ItemsSource = GetRealItems();
}
in MVVMLight ViewModels, this is shortcut-ed as
if (IsInDesignMode)
{
}
Similarly, since it looks like you're setting your ItemsSource in xaml, inside your class that is your DataContext, you could do something like
public class MyViewModel
{
public MyViewModel()
{
if (System.ComponentModel.DesignerProperties.IsInDesignTool)
{
Items = GenerateMockItems();
EditTime = GenerateRandomFutureDate();
}
else
{
//whatever you expect should happen in runtime
}
}
//what list is binding to
public ObservableCollection<Item> Items { get; set; }
//other properties.. for example
public bool HasItems { get { return Items != null && Items.Count > 0; } }
public DateTime EditDate { get; set; }
private ObservableCollection<Item> GenerateMockItems()
{
var collection = new ObservableCollection<Item>();
for (int i = 0; i < 10; i++)
{
collection.Add(new Item() { Name="sdfsdfs" , Channel=i });
}
return collection;
}
private DateTime GenerateRandomFutureDate()
{
return DateTime.Now.AddSeconds(new Random().Next(0,50000));
}
}