我想将数据网格中的按钮变为可见或折叠,这取决于本地窗口属性。
以下项目适用于 LayoutRoot
<navigation:Page xmlns:sdk="http://schemas.microsoft.com/winfx/2006/xaml/presentation/sdk" x:Class="TBPM.PageIssues"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
x:Name="PageIssueRoot"
...
_
<Grid DataContext="{Binding ElementName=PageIssueRoot}">
<Button Click="btnPasteMessage_Click" Visibility="{Binding Path=IsSaving, Converter={StaticResource BoolToVis}}" >
<Grid>
<Image Height="24" Source="/Next.png" HorizontalAlignment="Center" />
</Grid>
</Button>
</Grid>
代码隐藏
#Region "IsSaving"
''' <summary>
''' IsSaving Dependency Property
''' </summary>
Public Shared ReadOnly IsSavingProperty As DependencyProperty = _
DependencyProperty.Register("IsSaving", GetType(Boolean), GetType(PageIssues), _
New Windows.PropertyMetadata(False))
''' <summary>
''' Gets or sets the IsSaving property. This dependency property
''' indicates ....
''' </summary>
Public Property IsSaving() As Boolean
Get
Return CType(GetValue(IsSavingProperty), Boolean)
End Get
Set(ByVal value As Boolean)
SetValue(IsSavingProperty, value)
End Set
End Property
#End Region
如果将其用作 DataGrid 模板,则相同的代码将不起作用。
解决方案是什么?为什么模板找不到根窗口?