將控件綁定到由域對象提供的xaml

Question

我們有一個我們想要由用戶控制的UI區域。基本上他們爲我們提供了一塊xaml，它定義了特定區域的外觀。

public class Project 
{ 
    public string DisplaySpecificationXml { get; set; } 
} 

是否有從碰巧知道XAML，使我們可以在運行時看到它的域對象的屬性綁定一個簡單的方法;

PS - 請注意，所查看的項目將在運行時更改，然後我需要更新UI的這些區域。

Answer 1

<Window x:Class="MiscSamples.RuntimeXAML" 
     xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" 
     xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" 
     xmlns:local="clr-namespace:MiscSamples" 
     Title="RuntimeXAML" Height="300" Width="300"> 
    <Window.Resources> 
     <local:stringToUIConverter x:Key="Converter"/> 
    </Window.Resources> 
    <UniformGrid Rows="1" Columns="2"> 
     <ListBox ItemsSource="{Binding Projects}" x:Name="Lst"/> 
     <ContentPresenter Content="{Binding SelectedItem.DisplaySpecificationXml, ElementName=Lst, Converter={StaticResource Converter}}"/> 
    </UniformGrid> 
</Window> 

代碼背後：

public partial class RuntimeXAML : Window 
    { 
     public List<Project> Projects { get; set; } 

     public RuntimeXAML() 
     { 
      InitializeComponent(); 
      Projects = new List<Project> 
          { 
           new Project() 
            { 
             DisplaySpecificationXml = 
             "<StackPanel>" + 
             "<TextBlock FontWeight='Bold' Text='This is UserControl1'/>" + 
             "<ComboBox Text='ComboBox'/>" + 
             "</StackPanel>" 
            }, 
           new Project() 
            { 

            } 
          }; 
      DataContext = this; 
     } 
    } 

轉換器：

public class stringToUIConverter: IValueConverter 
    { 
     public object Convert(object value, Type targetType, object parameter, CultureInfo culture) 
     { 
      if (value == null || (!(value is string))) 
       return null; 

      var header = "<Grid xmlns='http://schemas.microsoft.com/winfx/2006/xaml/presentation' " + 
         "xmlns:x='http://schemas.microsoft.com/winfx/2006/xaml'>"; 

      var footer = "</Grid>"; 

      var xaml = header + (string) value + footer; 

      var UI = System.Windows.Markup.XamlReader.Parse(xaml) as UIElement; 
      return UI; 
     } 

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

結果：