StyleSelector和XAML的返回样式

Question

我在XAML中创建了一个样式,如何在样式选择器(代码)中返回此样式？

我在XAML中创建了样式,我想只返回在XAML中声明的样式.

Answer 1

您可以向您的属性添加属性StyleSelector,然后使用该属性将引用传递给StyleXAML.

public class MyStyleSelector : StyleSelector
{
    private Style styleToUse;

    public Style StyleToUse
    {
        get { return styleToUse; }
        set { styleToUse = value; }
    }

    public override Style SelectStyle(object item, DependencyObject container)
    {
        return styleToUse;
    }
}

<Control StyleSelector="{DynamicResource myStyleSelector}">
    <Control.Resources>
        <Style x:Key="myStyle">
        ...
        </Style>
        <local:MyStyleSelector x:Key="myStyleSelector" StyleToUse="{StaticResource myStyle}"/>
    </Control.Resources>
</Control>

Answer 2

您需要访问存储样式的XAML资源.通常,他们这样做的方法是将其存储在单独的资源文件中.然后,您需要访问该XAML文件的URI作为ResourceDictionary对象.这是一个示例,我使用转换器来决定元素将获得哪种样式.

namespace Shared.Converters
{
  public class SaveStatusConverter : IValueConverter
  {

    public object Convert(object value, System.Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {

      bool? saveState = (bool?)value;
      Uri resourceLocater = new Uri("/Shared;component/Styles.xaml", System.UriKind.Relative);
      ResourceDictionary resourceDictionary = (ResourceDictionary)Application.LoadComponent(resourceLocater);
      if (saveState == true)
        return resourceDictionary["GreenDot"] as Style;
      if (saveState == false)
        return resourceDictionary["RedDot"] as Style;
      return resourceDictionary["GrayDot"] as Style;

    }

    public object ConvertBack(object value, System.Type targetType, object parameter, System.Globalization.CultureInfo culture)
    {
      throw new System.NotImplementedException();
    }
  }
}