Trying to create Window type using FsXaml.Wpf v 3.1.6 using the XAML<"views\MyXaml.xaml"> syntax.
The Xaml file is in a directory called views within the project and has been made a "Resource" with Copy to Output Directory set to Do not copy.
I get the error Unable to load XAML data. Verify that all .xaml files are compiled as "Resource".
Repro steps
Create Xaml file MyXaml.xaml within a folder called views.
Set the Build Action to Resource. Copy to Output Directory to Do not copy
Create a type
type myWindow =XAML<"views\MyXaml.xaml">
Expected behavior
The type to be created successfully.
Actual behavior
Exception as mentioned above.
Known workarounds
Use __SOURCE_DIRECTORY__ in the dev environment and use XamlFileLocation but this fails with the executable.
Description
Trying to create Window type using FsXaml.Wpf v 3.1.6 using the XAML<"views\MyXaml.xaml"> syntax.
The Xaml file is in a directory called views within the project and has been made a "Resource" with Copy to Output Directory set to Do not copy. I get the error Unable to load XAML data. Verify that all .xaml files are compiled as "Resource".
Repro steps
Expected behavior
The type to be created successfully.
Actual behavior
Exception as mentioned above.
Known workarounds
Use __SOURCE_DIRECTORY__ in the dev environment and use XamlFileLocation but this fails with the executable.