Question

I am trying to use easyMock to write a test, that tests SecurityException in the following code.

eg. for NumberFormatException I use the below.

EasyMock.expect(mockEntityManager.find(UserProfile.class,”abc”)).andThrow(new
NumberFormatException());

Any ideas on what to expect to throw SecurityException?

public Object getAsObject(FacesContext facesContext, UIComponent
uiComponent, String s) {

EntityManager entityManager = (EntityManager)Component.getInstance(“entityManager”);

  if (s == null || s.equals("null")) {
        return null;      } else {
        try {
            long i = Long.parseLong(s);
            return entityManager.find(UserProfile.class, i);
        } catch (NumberFormatException e) {
            logger.error(e);
        } catch (SecurityException e) {
            logger.error(e);
        }         }

    return null;  }

Answer 1

Thanks for your responses..here is what I did to expect SecurityException.

MyClass abc = new MyClass();

EasyMock.expect(mockEntityManager.find(MyClass.class,111L)).andThrow(new SecurityException());

EasyMock.replay(mockEntityManager);

Object target = abc.getAsObject(mockFacesContext, mockUiComponent,"111");

Assert.assertEquals(null, target);