我有一个带有一些值的ComboBox,我想要同时处理两件事。
这是我的ComboBox,我想将10显示为默认值,并将其绑定到double?距离财产。
<ComboBox Grid.Row="5" Grid.Column="1"
SelectedIndex="1"
SelectedValue="{Binding Distance, Mode=TwoWay, Converter={StaticResource StringToDoubleConverter}}">
<ComboBoxItem>1</ComboBoxItem>
<ComboBoxItem IsSelected="True">10</ComboBoxItem>
<ComboBoxItem>100</ComboBoxItem>
<ComboBoxItem>1000</ComboBoxItem>
</ComboBox>
这是转换器:
public class StringToDoubleConverter : IValueConverter
{
public object Convert(object value, Type targetType, object parameter, string language)
{
return null;
}
public object ConvertBack(object value, Type targetType, object parameter, string language)
{
ComboBoxItem item = value as ComboBoxItem;
if (item != null)
{
double d;
if (double.TryParse(item.Content.ToString(), out d))
return d;
}
return null;
}
}
问题在于,在此代码中,所选项目10未在应用程序的开头显示。 如果我将删除转换器的行,那么它将显示所选的项目10,但是,我不能将它绑定到双精度?距离财产。我不想为它编写代码,例如:Convert.ToDouble(combobox1.SelectedValue)...
我可以做些什么来使两件事都有效?
答案 0 :(得分:0)
您需要从ViewModel填充组合框项目。此外,您不应使用SelectedValue
属性,而应使用SelectedItem
。请参阅下面给出的代码。
XAML
<ComboBox x:Name="cmb" ItemsSource="{Binding DistanceCollection}"
SelectedItem="{Binding Distance, Converter={StaticResource StringToDoubleConverter}, Mode=TwoWay}"/>
视图模型
public class viewModel : INotifyPropertyChanged
{
public viewModel()
{
DistanceCollection = new ObservableCollection<string>
{
"1",
"10",
"100",
"1000"
};
Distance = double.Parse(DistanceCollection[1].ToString());
}
public ObservableCollection<string> DistanceCollection { get; set; }
private double _Distance;
public double Distance
{
get
{
return _Distance;
}
set
{
_Distance = value;
OnPropertyChanged("Distance");
}
}
public event PropertyChangedEventHandler PropertyChanged;
protected void OnPropertyChanged(string propertyName = null)
{
var eventHandler = this.PropertyChanged;
if (eventHandler != null)
{
eventHandler(this, new PropertyChangedEventArgs(propertyName));
}
}
}
转换器
public class StringToDoubleConverter : IValueConverter
{
public object Convert(object value, Type targetType, object parameter, string language)
{
return value.ToString();
}
public object ConvertBack(object value, Type targetType, object parameter, string language)
{
string item = value as string;
if (!string.IsNullOrWhiteSpace(item))
{
double d;
if (double.TryParse(item, out d))
return d;
}
return null;
}
}