mockMvc POST Junit测试失败，并带有HttpMessageNotReadableException

Question

我正在尝试对我在控制器中创建的发布请求进行JUNIT测试。 尽管GET成功，但POST失败。

public class SampleObject implements Serializable {

    private static final long serialVersionUID = 1L;
    private String name;
    public SampleObject(String name) {
        this.name = name;
    }
    public String getName() {
        return name;
    }
}

样品要求

    @RequestMapping(value = "/post", method = RequestMethod.POST, produces="application/json", consumes="application/json")
    @ResponseBody
    public SampleObject postActiveSource(@RequestBody SampleObject inputObject) {
        return inputObject;
    }

这是我的模拟测试。

public class MockMvcHtmlUnitCreateMessageTest {


MockMvc mockMvc;

@InjectMocks
SampleController controller; 

 @Before
  public void setup() {
    MockitoAnnotations.initMocks(this);

    this.mockMvc = standaloneSetup(controller)
            .setMessageConverters(new MappingJackson2HttpMessageConverter()).build();
  }

@Test
public void test() {

    SampleObject so = controller.new SampleObject("abcd");
    ObjectMapper mapper = new ObjectMapper();


    try {
        this.mockMvc.perform(post("/service/sample/post").content(mapper.writeValueAsString(so)).contentType(MediaType.APPLICATION_JSON).accept(MediaType.APPLICATION_JSON)).andDo(print());

    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
 }

}

我得到这个例外。

Resolved Exception:
          Type = org.springframework.http.converter.HttpMessageNotReadableException

但是，获取请求的效果很好。 这是怎么了？ 任何帮助，将不胜感激。

编辑1：

我注意到以下异常。 这可能会有所帮助。

nested exception is com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class com..rest.controller.SampleController$SampleObject]: can not instantiate from JSON object (need to add/enable type information?)
 at [Source: org.springframework.mock.web.DelegatingServletInputStream@c792d4; line: 1, column: 2]

编辑2：

根据以下建议 ，我更改了构造函数，

public SampleObject(@JsonProperty("name") String name) {
            this.name = name;
}

但是，我得到以下异常

Caused by: java.lang.IllegalArgumentException: Argument #0 of constructor [constructor for SampleController$SampleObject, annotations: [null]] has no property name annotation; must have name when multiple-paramater constructor annotated as Creator at com.fasterxml.jackson.databind.deser.BasicDeserializerFactory.findValueInstantia‌​tor(BasicDeserializerFactory.java:287) ~[jackson-databind-2.3.4.jar:2.3.4] 

Answer 1

杰克逊默认情况下需要POJO类的无参数构造函数。 如果需要使用参数化的构造函数，则需要注释其参数。

public SampleObject(@JsonProperty("name") String name) {
    this.name = name;
}

Answer 2

您可能必须在配置文件中配置MessageConverter 。 如果使用Java配置：

 protected void configureMessageConverters(List<HttpMessageConverter<?>> converters) {
        ObjectMapper mapper = new ObjectMapper();
        mapper.setSerializationInclusion(Include.NON_NULL);

        MappingJackson2HttpMessageConverter jacksonConverter = new MappingJackson2HttpMessageConverter();
        jacksonConverter.setObjectMapper(mapper);

        converters.add(jacksonConverter);
    }