spring 通用异常处理,ajax异常返回json

直接看代码 ,实现spring HandlerExceptionResolver 类

这样子可以ajax请求返回json

spring-servlet.xml需要如下配置

<!-- 抛出controller页面或者 Jackson2JsonView -->

<bean class="com.hf.delaise.interceptor.ExceptionHandler" />

public class ExceptionHandler implements HandlerExceptionResolver {

Logger logger = LoggerFactory.getLogger(this.getClass());

@Override
public ModelAndView resolveException(HttpServletRequest request,
HttpServletResponse reponse, Object obj, Exception ex) {
String rqType = request.getHeader("X-Requested-With");//ajax的请求头
//String path = request.getServletPath(); //path
if (rqType != null && rqType.contains("XMLHttpRequest")) {
ModelAndView mav = new ModelAndView();
logger.error(ex.getMessage()+"=============="+ex);
MappingJackson2JsonView view = new MappingJackson2JsonView();
CallResult.returnError("系统异常,请联系客服!");
Map<String, Object> attributes = new HashMap<String, Object>();
attributes.put("message", "系统异常,请联系客服!");
attributes.put("success", false);
view.setAttributesMap(attributes);
mav.setView(view);
return mav;
} else {
logger.error(ex.getMessage()+"=============="+ex);
String model = "/error/error";
request.setAttribute("message", ex.getMessage());
return new ConsoleModelAndView(model,request);
}

}

}