use*_*925 2 c# wpf user-interface contextmenu popupmenu
以下类派生自System.Windows.Controls.UserControl.在所述类中,我调用OpenFileDialog打开XAML文件(工作流文件).接下来,我在右键单击鼠标时实现动态菜单.菜单没有显示.这是线程问题还是UI问题?在我的研究中,我一直无法找到解决方案.
提前致谢.
private void File_Open_Click(object sender, RoutedEventArgs e)
{
var fileDialog = new OpenFileDialog();
fileDialog.Title = "Open Workflow";
fileDialog.Filter = "Workflow| *.xaml";
if (fileDialog.ShowDialog() == DialogResult.OK)
{
LoadWorkflow(fileDialog.FileName);
MouseDown += new System.Windows.Input.MouseButtonEventHandler(mouseClickedResponse);
}
}
private void mouseClickedResponse(object sender, System.Windows.Input.MouseEventArgs e)
{
if (e.RightButton == MouseButtonState.Pressed)
{
LoadMenuItems();
}
}
private void LoadMenuItems()
{
System.Windows.Controls.ContextMenu contextmenu = new System.Windows.Controls.ContextMenu();
System.Windows.Controls.MenuItem item1 = new System.Windows.Controls.MenuItem();
item1.Header = "A new Test";
contextmenu.Items.Add(item1);
this.ContextMenu = contextmenu;
this.ContextMenu.Visibility = Visibility.Visible;
}
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