我在MVVM,WPF工作,我有一个弹出窗口;在这个弹出窗口中是一个列表框,在列表框中我有一个复选框。问题是:如果我从列表框中取消选中一个项目并单击“外部”,弹出窗口将消失;如果我再次点击,则复选框将重置为其初始值(所有项目都将被选中)。
那么,如何在应用程序运行时保持弹出窗口的状态并停止重置?我可以通过XAML做到这一点吗?
这是代码:
public class CheckedListItem<T> : INotifyPropertyChanged
{
public event PropertyChangedEventHandler PropertyChanged;
private bool isChecked = false;
private T item;
public CheckedListItem()
{ }
public CheckedListItem(T item, bool isChecked)
{
this.item = item;
this.isChecked = isChecked;
}
public T Item
{
get { return item; }
set
{
item = value;
if (PropertyChanged != null) PropertyChanged(this, new PropertyChangedEventArgs("Item"));
}
}
public bool IsChecked
{
get { return isChecked; }
set
{
isChecked = value;
if (PropertyChanged != null) PropertyChanged(this, new PropertyChangedEventArgs("IsChecked"));
}
}
}
viewModel:
private void OnApplyFiltersCommandRaised(object obj)
{
if (FilterElement.Contains("ClassView"))
{
switch (FilterElement)
{
case "buttonClassViewClassFilter":
FilteredClassViewItems.Clear();
FilteredFieldViewItems.Clear();
foreach (var filterItem in FilterItems)
{
if (filterItem.IsChecked == true)
{
FilteredClassViewItems.Add(classViewItems.First(c => c.ClassName == filterItem.Item));
FilteredFieldViewItems.Add(fieldViewItems.First(c => c.ClassName == filterItem.Item));
}
}
break;
...
public ObservableCollection<CheckedListItem<string>> FilterItems
{
get
{
return filterItems;
}
set
{
filterItems = value;
SetPropertyChanged("FilterItems");
}
}
XAML部分:
<ListBox x:Name="listBoxPopupContent"
Height="250"
ItemsSource="{Binding FilterItems}"
BorderThickness="0"
ScrollViewer.VerticalScrollBarVisibility="Auto">
<ListBox.Resources>
<Style TargetType="{x:Type ListBoxItem}">
<Setter Property="FontSize" Value="8" />
<Setter Property="IsSelected" Value="{Binding IsChecked, Mode=TwoWay}" />
</Style>
</ListBox.Resources>
<ListBox.ItemTemplate>
<DataTemplate>
<CheckBox IsChecked="{Binding IsChecked}"
Content="{Binding Item}"
Command="{Binding DataContext.ApplyFiltersCommand,
RelativeSource={RelativeSource FindAncestor,
AncestorType={x:Type ListBox}}}"
CommandParameter="{Binding IsChecked,
RelativeSource={RelativeSource Self},
Mode=OneWay}"/>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
提前致谢!
答案 0 :(得分:0)
如果要保持状态,可以创建一个包含列表框的新视图。然后你的弹出窗口将是
<Popup>
<views:MyListBoxview>
</Popup>
其中views是wpf可以找到MyListBoxview的路径。
这是一个如何做MyListBoxView的例子。首先,为项目添加一个新的usercontrol。然后你创建:
<ListBox ItemSource = {Binding MyCollectionOfItem}>
<ListBox.ItemTemplate>
<DataTemplate>
<CheckBox IsChecked = {Binding IsItemChecked} Content = {Binding Name}/>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
您需要为此视图分配一个视图模型,该视图模型当然会实现INotifyPropertyChanged并且将在其中定义这些类(此类也将实现INotifyPropertyChanged)
public class MyItem : INotifyPropertyChanged
{
public void SetPropertyChanged(string propertyName)
{
PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(propertyName));
}
public event PropertyChangedEventHandler PropertyChanged;
private bool isItemChecked = false;
public bool IsItemChecked
{
get { return isItemChecked; }
set
{
isItemChecked = value;
SetPropertyChanged("IsItemChecked");
}
}
private string name ;
public string Name
{
get { return Name; }
set
{
name = value;
SetPropertyChanged("Name");
}
}
}
最后,代表弹出窗口状态的viewmodel将包含在此属性
中private ObservableCollection<MyItem> myCollectionOfItem = new ObservableCollection<MyItem>();
public ObservableCollection<MyItem> MyCollectionOfItem
{
get { return myCollectionOfItem; }
set
{
myCollectionOfItem = value;
SetPropertyChanged("MyCollectionOfItem");
}
}
我通常会处理这种问题,正确地建模我需要绑定到WPF中的控件的对象