Ответ 1
В файле управления кодом элемента управления попробуйте:
this.Style = Resources["ResourceName"] as Style;
У меня есть следующий стиль, но мне нужно сделать его программным:
<xcdg:DataGridControl MinHeight="300"
Name="listViewUnallocated"
ItemsSource="{Binding Source={StaticResource
cvs_unallocatedTerminals}}"
AllowDrop="True"
Drop="Grid_Drop"
MouseMove="Grid_MouseMove"
KeyUp="listViewUnallocated_KeyUp"
MouseDoubleClick="gridUnallocated_MouseDoubleClick"
ReadOnly="True"
DockPanel.Dock="Top">
<xcdg:DataGridControl.Resources>
<Style TargetType="{x:Type xcdg:DataRow}" x:Name="selectedStyleTrigger">
<Style.Triggers>
<DataTrigger Binding="{Binding TerminalId}" Value="72948028">
<Setter Property="Background" Value="Red" />
</DataTrigger>
</Style.Triggers>
</Style>
</xcdg:DataGridControl.Resources>
В файле управления кодом элемента управления попробуйте:
this.Style = Resources["ResourceName"] as Style;
Установите x:Key
в XAML и в использовании кода:
something.Style = (Style) FindResource("YourResourceKey");
Привет, мы можем программировать стиль таким образом.
Style rowStyle = new Style(typeof(DataGridRow));
DataTrigger dataTrigger = new DataTrigger("TerminalId");
Binding binding = new Binding();
dataTrigger.Binding = binding;
dataTrigger.Value = 72948028;
Setter setter = new Setter(DataGridRow.BackgroundProperty, Brushes.Red);
dataTrigger.Setters.Add(setter);
rowStyle.Triggers.Add(dataTrigger);
listViewUnallocated.RowStyle = rowStyle;
Для изменения стиля кнопки
1-й: определить стили ресурса
<Window.Resources>
<Style x:Key="OvergroundIn" TargetType="Button">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="Button">
<Grid Background="#FF16832F">
<ContentPresenter TextBlock.Foreground="White" TextBlock.TextAlignment="Center" Margin="0,8,0,0" ></ContentPresenter>
</Grid>
</ControlTemplate>
</Setter.Value>
</Setter>
<Style.Triggers>
<Trigger Property="IsMouseOver" Value="True">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="Button">
<Grid Background="#FF06731F">
<ContentPresenter TextBlock.Foreground="White" TextBlock.TextAlignment="Center" Margin="0,8,0,0" ></ContentPresenter>
</Grid>
</ControlTemplate>
</Setter.Value>
</Setter>
</Trigger>
</Style.Triggers>
</Style>
<Style x:Key="OvergroundOut" TargetType="Button">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="Button">
<Grid Background="#FFF35E5E">
<ContentPresenter TextBlock.Foreground="White" TextBlock.TextAlignment="Center" Margin="0,8,0,0" ></ContentPresenter>
</Grid>
</ControlTemplate>
</Setter.Value>
</Setter>
<Style.Triggers>
<Trigger Property="IsMouseOver" Value="True">
<Setter Property="Template">
<Setter.Value>
<ControlTemplate TargetType="Button">
<Grid Background="#FFE34E4E">
<ContentPresenter TextBlock.Foreground="White" TextBlock.TextAlignment="Center" Margin="0,8,0,0" ></ContentPresenter>
</Grid>
</ControlTemplate>
</Setter.Value>
</Setter>
</Trigger>
</Style.Triggers>
</Style>
</Window.Resources>
2-й код кнопки определения
<Border Grid.Column="2" BorderBrush="Red" BorderThickness="2" CornerRadius="3" Margin="2,2,2,2" >
<Button Name="btnFichar" BorderThickness="0" Click="BtnFichar_Click">
<Button.Content>
<Grid Name="gridFichar">
<TextBlock Margin="0,7,0,7" TextAlignment="Center">Fichar</TextBlock>
</Grid>
</Button.Content>
</Button>
</Border>
3-й код позади
public void ShowStatus()
{
switch (((MainDto)this.DataContext).State)
{
case State.IN:
this.btnFichar.BorderBrush = new SolidColorBrush(Color.FromRgb(243, 94, 94));
this.btnFichar.Style = Resources["OvergroundIn"] as Style;
this.btnFichar.Content = "Fichar Salida";
break;
case State.OUT:
this.btnFichar.BorderBrush = new SolidColorBrush(Color.FromRgb(76, 106, 83));
this.btnFichar.Style = Resources["OvergroundOut"] as Style;
this.btnFichar.Content = "Fichar Entrada";
break;
}
}