WPF/MVVM: Sync scrolling of two datagrids in different views

Question

I have two datagrids side by side bound to different data tables and each with their own view.

The datatables both have the same number of rows, and I want both grids to

Answer 1

The best way I've used so far is to use the VisualTreeHelper class to find the correct ScrollViewer object (grid or no grid). I've used this in several projects.

Try this if any of you need it:

private static bool ScrollToOffset(DependencyObject n, double offset)
{
    bool terminate = false;
    for (int i = 0; i < VisualTreeHelper.GetChildrenCount(n); i++)
    {
        var child = VisualTreeHelper.GetChild(n, i);
        if (child is ScrollViewer)
        {
            (child as ScrollViewer).ScrollToVerticalOffset(offset);
            return true;
        }
    }
    if (!terminate)
        for (int i = 0; i < VisualTreeHelper.GetChildrenCount(n); i++)
            terminate = ScrollToOffset(VisualTreeHelper.GetChild(n, i), offset);
     return false;
}

Note: I typically use ListBox classes and would pass it directly to this function.

Happy programming :)

Answer 2

Take a look at codeproject Scroll Synchronization

Answer 3

The Scroll Synchronization project doesn't work for Datagrid because it doesn't expose ScrollToVerticalOffset

Answer 4

I was able to overcome this issue via some reflection hacks:

<DataGrid Name="DataGrid1" ScrollViewer.ScrollChanged="DataGrid1_ScrollChanged" />
<DataGrid Name="DataGrid2" />

and the code itself is:

    private void DataGrid1_ScrollChanged(object sender, ScrollChangedEventArgs e)
    {
        if (e.HorizontalChange != 0.0f)
        {
            ScrollViewer sv = null;
            Type t = DataGrid1.GetType();
            try
            {
                sv = t.InvokeMember("InternalScrollHost", BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.GetProperty, null, DataGrid2, null) as ScrollViewer;
                sv.ScrollToHorizontalOffset(e.HorizontalOffset);
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }
    }