Open WPF form when clicking hyperlink on WPF

Question

Hi, I want to open a new WPF form when I click the WPF hyperlink. How can I do it. I have seen many examples that open only the web url, but I want to open a new WPF form.

Answer 1

You can achieve this as follows:

<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 can simply 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 can 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, because once the dialog is closed, it cannot be reopened, which means that when you click the hyperlink again, an exception will be thrown.

---

Using interactivity, you can do this without such problems (link to the Blend SDK required):

 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> 

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

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(); } 

Answer 4

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 }