如何在Spring MVC中处理400错误

Question

当我传递无效的json格式时，我得到了400个Http响应，
我想返回自定义的json消息而不是这样，有人能建议在Spring 4.1中该怎么办吗？

使用ControllerAdvision处理执行，但它不起作用。

@ControllerAdvice
public class GlobalControllerExceptionHandler  {

       @ExceptionHandler({org.springframework.http.converter.HttpMessageNotReadableException.class})
       @ResponseStatus(HttpStatus.BAD_REQUEST)
            @ResponseBody
      public String resolveException() {
        return "error";
        }

}

<bean
        class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
        <property name="order" value="1" />
        <property name="mediaTypes">
            <map>
                <entry key="json" value="application/json" />
            </map>
        </property>
        <property name="defaultViews">
            <list>
                <!-- Renders JSON View -->
                <bean
                    class="org.springframework.web.servlet.view.json.MappingJacksonJsonView" />
            </list>
        </property>
    </bean>

Request 1: Empty json request : {}
Response Status Code: 400 Bad Request
Response Message    : Json request contains invalid data:null


Request 2:Invalid format of Json Request : {"data":,"name":"java"}
Response Status Code: 400 Bad Request
Response  or Exception message     :

nested exception is com.fasterxml.jackson.databind.JsonMappingException: Unexpected character (',' (code 44)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
 at [Source: com.ibm.ws.webcontainer.srt.http.HttpInputStream@8f308f3; line: 5, column: 57]

Answer 1

您可以尝试以这种方式映射异常。此代码将返回一个400状态，但您可以更改返回的方式与您发布的链接相同

@ExceptionHandler
@ResponseStatus(HttpStatus.BAD_REQUEST)
public void handleJsonMappingException(JsonMappingException ex) {}