单击WPF超链接时打开WPF表单

Question

嘿我想点击WPF超链接时打开一个新的WPF表单.我怎样才能做到这一点.我看过很多只打开网址的例子,但我想打开一个新的WPF表单.

Answer 1

你可以这样做:

<Label Height="25" Margin="26,27,116,0" Name="label1" VerticalAlignment="Top">
    <Hyperlink Click="Hyperlink_Click">Click Me</Hyperlink>
</Label>

并像这样处理它:

private void Hyperlink_Click(object sender, RoutedEventArgs e)
{
    Window2 form2 = new Window2();
    form2.Show();
}

Answer 2

你可以只处理click事件:

<Hyperlink Click="Hyperlink_Click">Link</Hyperlink>

private void Hyperlink_Click(object sender, RoutedEventArgs e)
{
    Dialogue diag = new Dialogue();
    diag.Show();
}

你也可以对XAML发疯:

<Hyperlink>
    <Hyperlink.Style>
        <Style TargetType="{x:Type Hyperlink}">
            <Style.Triggers>
                <EventTrigger RoutedEvent="Hyperlink.Click">
                    <BeginStoryboard>
                        <Storyboard>
                            <ObjectAnimationUsingKeyFrames Storyboard.TargetProperty="Visibility">
                                <Storyboard.Target>
                                    <local:Dialogue />
                                </Storyboard.Target>
                                <DiscreteObjectKeyFrame KeyTime="0:0:0" Value="{x:Static Visibility.Visible}"/>
                            </ObjectAnimationUsingKeyFrames>
                        </Storyboard>
                    </BeginStoryboard>
                </EventTrigger>
            </Style.Triggers>
        </Style>
    </Hyperlink.Style>
    <Hyperlink.Inlines>
        <Run Text="Open Dialogue"/>
    </Hyperlink.Inlines>
</Hyperlink>

然而,这是非常有问题的,因为一旦对话关闭它就无法重新打开,这意味着当您再次点击超链接时,将抛出异常.

使用交互性可以在没有这些问题的情况下执行此操作(但需要引用Blend SDK):

xmlns:i="clr-namespace:System.Windows.Interactivity;assembly=System.Windows.Interactivity"

<Hyperlink>
    <i:Interaction.Triggers>
        <i:EventTrigger EventName="Click">
            <t:CreateDialogAction Type="{x:Type local:Dialogue}"/>
        </i:EventTrigger>
    </i:Interaction.Triggers>
    <Hyperlink.Inlines>
        <Run Text="Open Dialogue"/>
    </Hyperlink.Inlines>
</Hyperlink>

对此的行动:

public class CreateDialogAction : TriggerAction<Hyperlink>
{
    public Type Type { get; set; }

    protected override void Invoke(object parameter)
    {
        if (Type != null && Type.IsSubclassOf(typeof(Window)) && Type.GetConstructor(Type.EmptyTypes) != null)
        {
            Window window = Type.GetConstructor(Type.EmptyTypes).Invoke(null) as Window;
            window.Show();
        }
    }
}