强制WPF在ItemsControl中创建项

如果ListBox使用VirtualzingStackPanel,只需快速查看,也许用StackPanel代替它就足够了

<ListBox.ItemsPanel>
  <ItemsPanelTemplate>
      <StackPanel/>
  <ItemsPanelTemplate>
<ListBox.ItemsPanel>

你可能走错了方向。我所做的是连接我的DataTemplate的[内容]的已加载事件:

<DataTemplate DataType="{x:Type local:ProjectPersona}">
  <Grid Loaded="Row_Loaded">
    <!-- ... -->
  </Grid>
</DataTemplate>

…然后处理事件处理程序中新显示的行:

private void Row_Loaded(object sender, RoutedEventArgs e)
{
    Grid grid = (Grid)sender;
    Carousel c = (Carousel)grid.FindName("carousel");
    ProjectPersona project = (ProjectPersona)grid.DataContext;
    if (project.SelectedTime != null)
        c.ScrollItemIntoView(project.SelectedTime);
}

这种方法在第一次显示行时对其进行初始化/检查,因此不会在前面执行所有行。如果你能接受这一点,那么也许这是更优雅的方法。

我想我知道怎么做了。问题是生成的项没有添加到可视化树中。经过一些搜索,我能想到的最好办法就是调用 VirtualizingStackPanel 在 ListBox . 虽然这并不理想,因为它只是为了测试,我想我不得不接受它。

这就是我的工作原理:

VirtualizingStackPanel itemsPanel = null;
FrameworkElementFactory factory = control.ItemsPanel.VisualTree;
if(null != factory)
{
    // This method traverses the visual tree, searching for a control of
    // the specified type and name.
    itemsPanel = FindNamedDescendantOfType(control,
        factory.Type, null) as VirtualizingStackPanel;
}

List<string> generatedItems = new List<string>();
IItemContainerGenerator generator = this.ItemsListBox.ItemContainerGenerator;
GeneratorPosition pos = generator.GeneratorPositionFromIndex(-1);
using(generator.StartAt(pos, GeneratorDirection.Forward))
{
    bool isNewlyRealized;
    for(int i = 0; i < this.ItemsListBox.Items.Count; i++)
    {
        isNewlyRealized = false;
        UIElement cntr = generator.GenerateNext(out isNewlyRealized) as UIElement;
        if(isNewlyRealized)
        {
            if(i >= itemsPanel.Children.Count)
            {
                itemsPanel.GetType().InvokeMember("AddInternalChild",
                    BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.InvokeMember,
                    Type.DefaultBinder, itemsPanel,
                    new object[] { cntr });
            }
            else
            {
                itemsPanel.GetType().InvokeMember("InsertInternalChild",
                    BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.InvokeMember,
                    Type.DefaultBinder, itemsPanel,
                    new object[] { i, cntr });
            }

            generator.PrepareItemContainer(cntr);
        }

        string itemText = GetControlText(cntr);
        generatedItems.Add(itemText);
    }
}

安迪的解决方案是一个很好的主意,但并不完整。例如,将在面板中创建前5个容器。名单上有300个>项目。我请求使用此逻辑的最后一个容器ADD。然后我请求最后一个index-1容器,带有这个logis ADD!这就是问题所在。面板内子级的顺序无效。

解决方案如下:

    private FrameworkElement GetContainerForIndex(int index)
    {
        if (ItemsControl == null)
        {
            return null;
        }

        var container = ItemsControl.ItemContainerGenerator.ContainerFromIndex(index -1);
        if (container != null && container != DependencyProperty.UnsetValue)
        {
            return container as FrameworkElement;
        }
        else
        {

            var virtualizingPanel = FindVisualChild<VirtualizingPanel>(ItemsControl);
            if (virtualizingPanel == null)
            {
                // do something to load the (perhaps currently unloaded panel) once
            }
            virtualizingPanel = FindVisualChild<VirtualizingPanel>(ItemsControl);

            IItemContainerGenerator generator = ItemsControl.ItemContainerGenerator;
            using (generator.StartAt(generator.GeneratorPositionFromIndex(index), GeneratorDirection.Forward))
            {
                bool isNewlyRealized = false;
                container = generator.GenerateNext(out isNewlyRealized);
                if (isNewlyRealized)
                {
                    generator.PrepareItemContainer(container);
                    bool insert = false;
                    int pos = 0;
                    for (pos = virtualizingPanel.Children.Count - 1; pos >= 0; pos--)
                    {
                        var idx = ItemsControl.ItemContainerGenerator.IndexFromContainer(virtualizingPanel.Children[pos]);
                        if (!insert && idx < index)
                        {
                            ////Add
                            virtualizingPanel.GetType().InvokeMember("AddInternalChild", BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.InvokeMethod, Type.DefaultBinder, virtualizingPanel, new object[] { container });
                            break;
                        }
                        else
                        {
                            insert = true;
                            if (insert && idx < index)
                            {
                                break;
                            }
                        }
                    }

                    if (insert)
                    {
                        virtualizingPanel.GetType().InvokeMember("InsertInternalChild", BindingFlags.NonPublic | BindingFlags.Instance | BindingFlags.InvokeMethod, Type.DefaultBinder, virtualizingPanel, new object[] { pos + 1, container });
                    }
                }

                return container as FrameworkElement;
            }
        }
    }

对于任何想知道这一点的人来说,在Andy的例子中,也许用普通StackPanel替换VirtualzingStackPanel是最好的解决方案。

在ItemContainerGenerator上调用PrepareItemContainer无法工作的原因是,项目必须位于可视树中,PrepareItemContainer才能工作。对于VirtualzingStackPanel,在VirtualzingStackPanel确定该项目正在/即将出现在屏幕上之前,该项目不会被设置为面板的可视子项。

就我而言,我发现 UpdateLayout() ItemsControl ( ListBox , ListView )启动了它的 ItemContainerGenerator ,使生成器的状态从“NotStarted”更改为“GeneratingContainers”,以及 null 集装箱已不再由海关归还 ItemContainerGenerator.ContainerFromItem 和/或 ItemContainerGenerator.ContainerFromIndex

例如:

    public static bool FocusSelectedItem(this ListBox listbox)
    {
        int ix;
        if ((ix = listbox.SelectedIndex) < 0)
            return false;

        var icg = listbox.ItemContainerGenerator;
        if (icg.Status == GeneratorStatus.NotStarted)
            listbox.UpdateLayout();

        var el = (UIElement)icg.ContainerFromIndex(ix);
        if (el == null)
            return false;

        listbox.ScrollIntoView(el);

        return el == Keyboard.Focus(el);
    }