WPF中的动态菜单

时间:2013-10-10 14:45:33

标签: c# wpf

如何使用Bindings和Observable集合从具有Bindings和Observable集合的数据库表中将菜单项添加到WPF中的菜单控件(而不是上下文菜单)?我有这个菜单:

<Menu HorizontalAlignment="Left" Height="27" VerticalAlignment="Top" Width="649">
    <MenuItem Header="_File">
       <MenuItem Header="_Exit" Command="{Binding ExitCommand}"/>
       </MenuItem>
    <MenuItem Header="_MyMenu">
       <MenuItem Header="_SubMenu1" Command="{Binding  SubMenu1Command}" />
       <MenuItem Header="_SubMenu2" Command="{Binding  SubMenu2Command}" />
    </MenuItem>
</Menu>

“SubMenu1”和“_SuMenu2”是数据库表中的值:

codSubMenu | SubMenuColum | CommandColumn

1_的 _ __ _ __ _ __ | SubMenu1 _ __ _ _ | SubMenu1Command 2 _ __ _ __ _ ___ | SubMenu2 _ ____ | _SubMenu2Command

我需要这样的东西:

<Menu HorizontalAlignment="Left" Height="27" VerticalAlignment="Top" Width="649"
    ItemsSource="{Binding ObservableCollectionMenu}">
    <MenuItem Header="_File">
       <MenuItem Header="_Exit" Command="{Binding ExitCommand}"/>
    </MenuItem>
    <MenuItem Header="_MyMenu">
        <MenuItem Header="{Binding  ObservableCollectionMenu.SubMenuColumn}" Command="{Binding  ObservableCollectionMenu.CommandColumn}" />
    </MenuItem>
</Menu>

当我运行应用程序时,当我按下File和MyMenu选项时,菜单必须显示此内容:

文件| MyMenu菜单

退出| SubMenu1

___ | SubMenu2

3 个答案:

答案 0 :(得分:5)

使用菜单的ItemsSource属性和MenuItems(在样式中)来绑定您的集合:

<Menu ItemsSource="{Binding YourCollection}" />

<Style TargetType="MenuItem">
    <Setter Property="Header" Value="{Binding Path=Name}" />
    <Setter Property="ItemsSource" Value="{Binding Path=Children}" />
</Style>

编辑:对于命令绑定,请执行以下操作:

  1. 将这样的setter添加到MenuItem的模板中:

    <Setter Property="Command" Value="{Binding Path=Command}" />
    
  2. 将此结构用于MenuItem视图模型:

    public class BindableMenuItem
    {
         public string Name { get; set; }
         public BindableMenuItem[] Children { get; set; }
         public ICommand Command { get; set; }
    }
    
  3. 将根项添加到BindableMenuItems的集合中,并将此集合绑定到菜单中。

答案 1 :(得分:2)

这就是我解决它的方式,

我创建了一个MenuItem类(注意它有一个Items列表,所以你可以构建子菜单):

public class MenuItem : ModelBase<MenuItem>
{
    private List<MenuItem> _Items;

    public MenuItem(string header, ICommand command)
    {
        Header = header;
        Command = command;
    }

    public MenuItem()
    {

    }

    public string Header { get; set; }

    public List<IMenuItem> Items
    {
        get { return _Items ?? (_Items = new List<IMenuItem>()); }
        set { _Items = value; }
    }

    public ICommand Command { get; set; }
    public string CommandName { get; set; }
    public object Icon { get; set; }
    public bool IsCheckable { get; set; }
    private bool _IsChecked;
    public bool IsChecked
    {
        get { return _IsChecked; }
        set
        {
            _IsChecked = value;
            NotifyPropertyChanged(m=>m.IsChecked);
        }
    }

    public bool Visible { get; set; }
    public bool IsSeparator { get; set; }
    public string InputGestureText { get; set; }
    public string ToolTip { get; set; }
    public int MenuHierarchyID { get; set; }
    public int ParentMenuHierarchyID { get; set; }
    public string IconPath { get; set; }
    public bool IsAdminOnly { get; set; }
    public object Context { get; set; }
    public IMenuItem Parent { get; set; }
    public int int_Sequence { get; set; }
    public int int_KeyIndex { get; set; }
}

观点:

<Menu DockPanel.Dock="Top" ItemsSource="{Binding Path=MainMenu}">
    <Menu.ItemContainerStyle>
        <Style>
            <Setter Property="MenuItem.Header" Value="{Binding Path=Header}" />
            <Setter Property="MenuItem.ItemsSource" Value="{Binding Path=Items}" />
            <Setter Property="MenuItem.Icon" Value="{Binding Path=Icon}" />
            <Setter Property="MenuItem.IsCheckable" Value="{Binding Path=IsCheckable}" />
            <Setter Property="MenuItem.IsChecked" Value="{Binding Path=IsChecked,Mode=TwoWay,UpdateSourceTrigger=PropertyChanged}" />
            <Setter Property="MenuItem.Command" Value="{Binding Path=Command}" />
            <!--<Setter Property="MenuItem.CommandParameter" Value="{Binding Path=IsChecked}"/>-->
            <Setter Property="MenuItem.CommandParameter" Value="{Binding Path=.}"/>
            <Setter Property="MenuItem.InputGestureText" Value="{Binding Path=InputGestureText}"/>
            <Setter Property="MenuItem.ToolTip" Value="{Binding Path=ToolTip}" />
            <Style.Triggers>
                <DataTrigger Binding="{Binding Path=IsSeparator}" Value="true">
                    <Setter Property="MenuItem.Template">
                        <Setter.Value>
                            <ControlTemplate TargetType="{x:Type MenuItem}">
                                <Separator Style="{DynamicResource {x:Static MenuItem.SeparatorStyleKey}}" />
                            </ControlTemplate>
                        </Setter.Value>
                    </Setter>
                </DataTrigger>
            </Style.Triggers>
        </Style>
    </Menu.ItemContainerStyle>
</Menu>

MainMenu是我的主ViewModel中的ObservableCollection属性,您可以从数据库中填充该属性。

    public ObservableCollection<MenuItem> MainMenu
    {
        get { return _MainMenu; }
        set
        {
            _MainMenu = value;
            NotifyPropertyChanged(x => x.MainMenu);
        }
    }

答案 2 :(得分:0)

我在XAML中没有快速解决方案。我需要从数据库中获取子菜单项,根据具体的配置文件,一些用户拥有其他只有2或3项的所有项目。独特的方法是在XAML中创建具有禁用项的菜单,将菜单引用传递给ViewModel(如果是MVVM App)并与ObservableCollection进行比较,只启用项目equals:

<menu horizontalalignment="Left" height="27" verticalalignment="Top" width="649" name="menu1">
      <menuitem header="_File">
          <menuitem header="_Exit" command="{Binding ExitCommand}" />
      </menuitem>
      <menuitem header="_MyMenu">
          <menuitem header="_SubMenu1" command="{Binding  Command1}" isenabled="False" />
          <menuitem header="_SubMenu2" command="{Binding  Command2}" isenabled="False" />
      </menuitem>
</menu>

视图模型:

for (int i = 0; i < ObservableCollectionMenu.Count; i++)
{

    for (int j = 0; j < ((MenuItem)menu1.Items[1]).Items.Count; j++)
    {
         if (((MenuItem)((MenuItem)menu1.Items[1]).Items[j]).Header.ToString().Equals(ObservableCollectionMenu[i].SubMenuColumn))
         {
            ((MenuItem)((MenuItem)menu1.Items[1]).Items[j]).IsEnabled = true;
             break;
         }
    }
}

感谢所有回答我问题的人,stackoverflow可以更好地帮助代码项目。