Open WPF form when clicking WPF hyperlink

Question

I want to open a new WPF form when I click in a WPF hyperlink. I've seen a lot of examples that only opens web url, but I want to open a new WPF form.

Answer 1

You can achive this like this:

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

and handle it like this:

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

Answer 2

You could just handle the click event:

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

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

---

You could also go crazy with XAML:

<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>

This however is very problematic since once the dialogue is closed it cannot be reopened, that means when you click the hypelink again an exception will be thrown.

---

Using interactivity you could do this without such problems (needs a reference to the Blend SDK though):

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>

The action for this:

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();
        }
    }
}

Answer 3

And using MVVM you can do in your View

<Hyperlink NavigateUri="{Binding MyUri}" 
           Command="{Binding OpenHyperlinkCommand}">Link text
</Hyperlink>

and in your ViewModel

private ICommand _openHyperlinkCommand;
public ICommand OpenHyperlinkCommand {
    get
    {
        if (_openHyperlinkCommand == null) 
            _openHyperlinkCommand = new RelayCommand<object>(p => ExecuteHyperlink());
        return _openHyperlinkCommand;
    }
}

private void ExecuteHyperlink() {
    //do stuff here
}

Answer 4

XAML :

<TextBlock Height="23" HorizontalAlignment="Left" Margin="254,130,0,0" Name="textBlock1" VerticalAlignment="Top">
        <Hyperlink Click="Hyperlink_Click">Clique Aqui</Hyperlink>
        </TextBlock>

CodeBehind :

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

this?