我有一个Commobox在ViewModel中完美显示其中的列表,但是我希望拥有它,以便当选择列表中的选定项目时,它会触发ViewModel屏幕,我只想从列表中做到这一点?因此,这是我在选择视图中所拥有的:
<ComboBox x:Name="CatalogName1" SelectedItem="{Binding SelectedCatalog1}" Style="{DynamicResource appComboBox}" Grid.Row="1" Grid.Column="1" >
</ComboBox>
和selectViewModel:
public List<string> CatalogName1
{
get
{
return new List<string> { "New", "Replace", "Extended", "Nothing", "ShowScreen" };
}
}
private string selectedCatalog1;
public string SelectedCatalog1
{
get
{
return this.selectedCatalog1;
}
set
{
this.selectedCatalog1 = value;
this.NotifyOfPropertyChange(() => this.SelectedCatalog1);
}
}
组合列表中的" showcreen"应该显示showcreenviewmodel,但我已经尝试使用getter setter,这对我没有意义
好吧,这是我解决问题的方法...
private string selectedCatalog1;
public string SelectedCatalog1
{
get
{
return selectedCatalog1;
}
set
{
selectedCatalog1 = value;
ValidateValue(value);
NotifyOfPropertyChange(() => SelectedCatalog1);
}
}
private void ValidateValue(string s)
{
if (s == "ShowScreen")
{
ActivateItem(new ShowScreenViewModel());
}
}