How to display row numbers in a ListView?

Question

The obvious solution would be to have a row number property on a ModelView element, but the drawback is that you have to re-generate those when you add records or change sort order.

Is there an elegant solution?

Answer 1

I think you have the elegant solution, but this works.

XAML:

<ListView Name="listviewNames">
  <ListView.View>
    <GridView>
      <GridView.Columns>
        <GridViewColumn
          Header="Number"
          DisplayMemberBinding="{Binding RelativeSource={RelativeSource FindAncestor, 
                                         AncestorType={x:Type ListViewItem}}, 
                                         Converter={StaticResource IndexConverter}}" />
        <GridViewColumn
          Header="Name"
          DisplayMemberBinding="{Binding Path=Name}" />
      </GridView.Columns>
    </GridView>
  </ListView.View>
</ListView>

ValueConverter:

public class IndexConverter : IValueConverter
{
    public object Convert(object value, Type TargetType, object parameter, CultureInfo culture)
    {
        ListViewItem item = (ListViewItem) value;
        ListView listView = ItemsControl.ItemsControlFromItemContainer(item) as ListView;
        int index = listView.ItemContainerGenerator.IndexFromContainer(item);
        return index.ToString();
    }

    public object ConvertBack(object value, Type targetType, object parameter, CultureInfo culture)
    {
        throw new NotImplementedException();
    }
}

Answer 2

If you have a dynamic list where items are added, deleted or moved, you can still use this very nice solution and simply let the currentview of your listview refresh itself after the changements in your source list are done. This code sample removes the current item directly in the data source list "mySourceList" (which is in my case an ObservableCollection) and finally updates the line numbers to correct values .

ICollectionView cv = CollectionViewSource.GetDefaultView(listviewNames.ItemsSource);
if (listviewNames.Items.CurrentItem != null)
{
    mySourceList.RemoveAt(cv.CurrentPosition);
    cv.Refresh();
}

Answer 3

First you need to set the AlternationCount to items count+1, for instance:

<ListView AlternationCount="1000" .... />

Then AlternationIndex will show the real index, even during the scrolling:

 <GridViewColumn
       Header="#" Width="30"
       DisplayMemberBinding="{Binding (ItemsControl.AlternationIndex),
       RelativeSource={RelativeSource AncestorType=ListViewItem}}" />

Answer 4

This will work like a charm, I don't know about performance, Still we can give it a try

Create a Multi Value Converter

public class NumberingConvertor : IMultiValueConverter
 {
  public object Convert(object[] values, Type targetType, object parameter, System.Globalization.CultureInfo culture)
  {
   if (values != null && values.Any() && values[0] != null && values[1] != null)
   {
    //return (char)(((List<object>)values[1]).IndexOf(values[0]) + 97);
    return ((List<object>)values[1]).IndexOf(values[0]) + 1;
   }
   return "0";
  }

  public object[] ConvertBack(object value, Type[] targetTypes, object parameter, System.Globalization.CultureInfo culture)
  {
   return null;
  }
 }
}

and your Xaml like this

<ItemsControl ItemsSource="{Binding ListObjType}">
        <ItemsControl.ItemsPanel>
            <ItemsPanelTemplate>
                <StackPanel />
            </ItemsPanelTemplate>
        </ItemsControl.ItemsPanel>
        <ItemsControl.ItemTemplate>
            <DataTemplate>
                <StackPanel Orientation="Horizontal">
                    <Label>
                        <MultiBinding Converter="{StaticResource NumberingConvertor}">
                            <Binding Path="" />
                            <Binding Path="ItemsSource"
                                     RelativeSource="{RelativeSource AncestorType=ItemsControl}" />
                        </MultiBinding>
                    </Label>
                    <TextBlock Text="{Binding }" />
                </StackPanel>
            </DataTemplate>
        </ItemsControl.ItemTemplate>
    </ItemsControl>

Idea is to send Object and list both to the converter and let converter decide the number. You can modify converter to display ordered list.

Answer 5

Here is another way, including code comments that will help you understand how it works.

public class Person
{
    private string name;
    private int age;
    //Public Properties ....
}

public partial class MainWindow : Window
{

    List<Person> personList;
    public MainWindow()
    {
        InitializeComponent();

        personList= new List<Person>();
        personList.Add(new Person() { Name= "Adam", Agen= 25});
        personList.Add(new Person() { Name= "Peter", Agen= 20});

        lstvwPerson.ItemsSource = personList;
//After updates to the list use lstvwPerson.Items.Refresh();
    }
}

The XML

            <GridViewColumn Header="Number" Width="50" 
                DisplayMemberBinding="{ 
                    Binding RelativeSource= {RelativeSource Mode=FindAncestor, AncestorType={x:Type ListViewItem}},
                   DELETE Path=Content, DELETE
                    Converter={StaticResource IndexConverter}, 
                    ConverterParameter=1
                }"/>

RelativeSource is used in particular binding cases when we try to bind a property of a given object to another property of the object itself [1].

Using Mode=FindAncestor we can traverse the hierarchy layers and get a specified element, for example the ListViewItem (we could even grab the GridViewColumn). If you have two ListViewItem elements you can specify which you want with "AncestorLevel = x".

Path: Here I simply take the content of the ListViewItem (which is my object "Person").

Converter Since I want to display row numbers in my Number column and not the object Person I need to create a Converter class which can somehow transform my Person object to a corresponding number row. But its not possible, I just wanted to show that the Path goes to the converter. Deleting the Path will send the ListViewItem to the Converter.

ConverterParameter Specify a parameter you want to pass to the IValueConverter class. Here you can send the state if you want the row number to start at 0,1,100 or whatever.

public class IndexConverter : IValueConverter
{
    public object Convert(object value, Type TargetType, object parameter, System.Globalization.CultureInfo culture)
    {
        //Get the ListViewItem from Value remember we deleted Path, so the value is an object of ListViewItem and not Person
        ListViewItem lvi = (ListViewItem)value;
        //Get lvi's container (listview)
        var listView = ItemsControl.ItemsControlFromItemContainer(lvi) as ListView;

        //Find out the position for the Person obj in the ListView
//we can get the Person object from lvi.Content
        // Of course you can do as in the accepted answer instead!
        // I just think this is easier to understand for a beginner.
        int index = listView.Items.IndexOf(lvi.Content);

        //Convert your XML parameter value of 1 to an int.
        int startingIndex = System.Convert.ToInt32(parameter);

        return index + startingIndex;
    }

    public object ConvertBack(object value, Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {
        throw new NotImplementedException();
    }
}