Question

我有一个简单的验证来验证客户端是否存在具有相同属性，验证只有2个选项： 1.如果满足某些条件，则抛出验证异常。 2.否则返回无效。

像这样：

public class ClientValidator {
   private ClientFinder clientFinder;
   protected void validate(final Client client) throws ValidationException
   {
        final Client existingClient = clientFinder.getClientByAttributes(client.getAttr1(),
        client.getAttr2());
        if (existingClient != null && ! existingClient.getId().equals(client.getId()))
        {
            throw new ValidationException("Client Exists with the same attributes"));
        }
   }
   //getters and setters

}    
}

The Junit：

@UnitTest
@RunWith(MockitoJUnitRunner.class)
public class ValidatorTest
{
    private ClientValidator validator;

    @Mock
    private ClientFinder clientFinder;
    private Client existingClient;
    private Client newClient;
    private static final String ATTR1 = "ATTR1";
    private static final String ATTR2 = "ATTR2";

    @Before
    public void setup()
    {
        existingClient = new Client("ID1", ATTR1, ATTR2);
        newClient = new Client("ID2", ATTR1, ATTR2);


        validator = new ClientValidator();
        validator.setClientFinder(clientFinder);
    }

    @Test(expected = ValidatorException.class)
    public void testCreatingNewClient() throws ValidationException
    {
        //the service will find other Client than the one being created
        Mockito.when(clientFinder.getClientByAttributes(ATTR1, ATTR2)).thenReturn(existingClient);
        validator.onValidate(newClient);
    }



    @Test()
    public void testModifyExistingClient() throws ValidationException
    {
        //the service will find the same Client being modified
        Mockito.when(clientFinder.getClientByAttributes(ATTR1, ATTR2)).thenReturn(newClient);
        validator.onValidate(newClient); //No exception should be thrown
    }


}

第二次验证是可能发生的情况之一，是否由于没有断言而需要进行第二次验证？有没有办法验证第二个条件？

Answer 1

你写的测试很好。不需要“断言”。执行到达测试底部的事实足以使其“绿色”。你已经在测试中对这个效果做了评论，并且一定要把它放进去，只是为了清楚说明发生了什么。但请不要在测试中添加无意义的“断言”。

Junit用于void验证方法

1 个答案: