Set a Page for Component Testing with Apache Wicket

Question

I am testing some components in my wicket 7 application.

My Component is nothing special but it inherits from

public PageAwarePanel extends Panel {
    @Override
    protected void onInitialize() {
        super.onInitialize();

        //refuse if used on page without PageConfig
        if (getPageConfigurationModel() == null){
            throw new RuntimeException("this component is only allowed inside pages having PageConfigurationModel");
        }
    }

    protected IModel<PageConfiguration> getPageConfigurationModel(){
        if (getPage() instanceof TemplatePage){
            return ((TemplatePage)getPage()).getPageConfigurationModel();
        }               
        return null;
    } 
}

With this I can access some configurations from a certain panel.

Now when I try in a test:

    PositionsPanel p = new PositionsPanel("123", asmNumber, Model.of());
    tester.startPage(MyPage.class);
    tester.startComponentInPage(p);

where MyPage is a TemplatePage.

I get the defined RuntimeException. My Question is:

How can I test this component with defining on which page it should be rendered?

Thanks for all the help in advanced.

score 2 · Accepted Answer · answered Feb 24 '16 at 13:54

tester.startPage(MyPage.class);
tester.startComponentInPage(p);

those two are not related. It is like navigating two different pages in the browser.

The best way is to create a page for the tests that fulfills the requirements, add the panel to this page and do tester.startPage(TestPage.class).

Another way is to override org.apache.wicket.util.tester.BaseWicketTester#createPage() and return an instance of MyPage. This way you can still use startComponentInPage() but otherwise it is basically the same as the first approach.

Thank you martin, the first described approach works for me. — kism3t, Feb 25 '16 at 07:26

Set a Page for Component Testing with Apache Wicket

1 Answers1