我正在尝试对我在控制器中创建的帖子请求进行JUNIT测试。虽然GET成功,但POST不会。
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;
}
}
样品张贴请求
@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();
}
}
}
我得到了这个例外。
Resolved Exception:
Type = org.springframework.http.converter.HttpMessageNotReadableException
但是,get请求正常。这里出了什么问题?任何帮助将不胜感激。
EDIT1:
我注意到以下异常。这可能有所帮助。
nested exception is com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class com..rest.controller.SampleController$SampleObject]: can not instantiate from JSON object (need to add/enable type information?)
at [Source: org.springframework.mock.web.DelegatingServletInputStream@c792d4; line: 1, column: 2]
EDIT2:
基于a suggestion below,我改变了修改构造函数,
public SampleObject(@JsonProperty("name") String name) {
this.name = name;
}
然而,我得到以下异常
Caused by: java.lang.IllegalArgumentException: Argument #0 of constructor [constructor for SampleController$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.findValueInstantiator(BasicDeserializerFactory.java:287) ~[jackson-databind-2.3.4.jar:2.3.4]
答案 0 :(得分:2)
杰克逊希望默认情况下为您的POJO类提供无参数构造函数。如果需要参数化构造函数,则需要注释其参数。
public SampleObject(@JsonProperty("name") String name) {
this.name = name;
}
答案 1 :(得分:0)
您可能必须在配置文件中配置MessageConverter
。如果您使用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);
}