Mockito模拟方法，但使用其参数模拟返回

Question

有一个方法公共内容创建儿童（字符串路径，字符串内容类型，地图

我想用任何类型的参数来模拟该方法，因此when（）将不起作用，因为我必须告诉它该方法应该接收哪些参数才能真正模拟。

因此，我想对任何方法调用做出实际反应，而不依赖于它的给定参数（使用spies？）然后调用某种类型的“回调”来返回一个Content对象，我想用给该方法的实际参数来构建它。

我在Mockito中找不到支持此功能的特定API。




Answer 1

你可以用匹配器

MyClass m = mock(MyClass.class);
when(m.createChild(any(String.class), any(String.class), any(Map.class))).thenReturn(new Content());

您还应该能够以这种方式使用参数

when(m.createChild(any(String.class), any(String.class), any(Map.class))).thenAnswer(
       new Answer<Content>()
       {
           @Override
           public Content answer(final InvocationOnMock invocation) throws Throwable
           {
               return new Content((String) invocation.getArguments()[0], 
(String) invocation.getArguments()[1],
(Map) invocation.getArguments()[2]);
               }
           }
       }
       );

Answer 2

当然可以。我已经为此编写了简单的单元测试

public class MockitoTest {


    private SampleService sampleService;

    @Before
    public void setUp() throws Exception {
        sampleService = Mockito.mock(SampleService.class);
    }

    @Test
    public void mockitoTest() throws Exception {
        when(sampleService.createChild(anyString(), anyString(), anyMapOf(String.class, Object.class)))
            .thenAnswer(invocationOnMock -> {
                //Here you can build result based on params
                String pathArg = (String) invocationOnMock.getArguments()[0];
                if (pathArg.equals("p1")) {
                    return new Content("Content for p1");
                } else if (pathArg.equals("p2")) {
                    return new Content("Content for p2");
                } else {
                    return invocationOnMock.callRealMethod();
                }
            });

        Content content = sampleService.createChild("p1", "any", new HashMap<>());
        assertEquals("Content for p1", content.getData());

        content = sampleService.createChild("p2", "any", new HashMap<>());
        assertEquals("Content for p2", content.getData());

        content = sampleService.createChild("whatever", "any", new HashMap<>());
        assertEquals("original", content.getData());

    }

    /**
     * Sample service with method
     */
    private static class SampleService {

        public Content createChild(String path, String contentType, Map<String, Object> properties) {
            return new Content("original");
        }

    }

    /**
     * Content example
     */
    private static class Content {

        private String data;

        Content(String data) {
            this.data = data;
        }

        String getData() {
            return data;
        }
    }

}

Answer 3

您可以使用匹配器：

您可以尝试以下方法：

when(service.createChild(anyString(), anyString(), anyMap()))
     .thenReturn(any(Content.class));