Question

我在Code Behind中有自定义属性，我想在XAML中设置。

属性：

Public Property WindowName() As String Implements IVendorEntryEditView.WindowName
    Get
        Return CType(GetValue(WindowNameProperty), Integer)
    End Get
    Set(ByVal value As String)
        SetValue(WindowNameProperty, value)
    End Set
End Property

Public Shared ReadOnly WindowNameProperty As DependencyProperty = _
    DependencyProperty.Register("WindowName", GetType(String), GetType(VendorEntryEditView), _
        New PropertyMetadata(""))

但是，当我尝试设置时，我在XAML中收到Attached property has no setter消息：

<UserControl x:Class="VendorEntryEditView"
         xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
         xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
         xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
         xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
         xmlns:local="clr-namespace:EntryEditUi"
         mc:Ignorable="d"
         Loaded="VendorEntryEditView_OnLoaded"
         local:VendorEntryEditView.WindowName="test"
         d:DesignHeight="300" d:DesignWidth="300">

如何在XAML中设置该属性？谢谢！

Answer 1

也许只是尝试（使用UserControl的地方）：

<local:VendorEntryEditView WindowName="Test Value" />

或如果您必须在UserControl本身的xaml中指定此值，请通过Style

中的VendorEntryEditView.xaml进行分配

<UserControl.Style>
  <Style>
    <Setter Property="local:VendorEntryEditView.WindowName"
            Value="Blah Name" />
  </Style>
</UserControl.Style>

在XAML中设置UserControl属性

1 个答案: