web项目中处理捕获异常统一处理

 1 public class ExceptionInterceptor implements Interceptor { 2 ?3 ????//@Override 4 ????public void intercept(Invocation ai) { 5 ????????BaseController controller = (BaseController) ai.getController(); 6 ????????HttpServletRequest request = controller.getRequest(); 7 ????????boolean successed = false; 8 ????????String exception = ""; 9 ????????try {10 ????????????ai.invoke();11 ????????????successed = true;12 ????????} catch (Exception e) {13 ????????????e.printStackTrace();14 ????????????successed = false;15 ????????????// 判断是否ajax请求16 ????????????String header = request.getHeader("X-Requested-With");17 ????????????boolean isAjax = "XMLHttpRequest".equalsIgnoreCase(header);18 ????????????String msg = e.getMessage();19 ????????????exception = Exceptions.getStackTraceAsString(e);20 ????????????controller.keepPara();21 ????????????if (isAjax) {22 ????????????????msg = new StringBuilder().append("{\"source\":\"ExceptionInterceptor\",\"code\":\"1\",\"message\":\"系统异常：").append(msg).append("\"}").toString();23 ????????????????controller.renderJson(msg);24 ????????????} else {25 ????????????????String redirctUrl = request.getHeader("referer");26 ????????????????if (StringUtils.isBlank(redirctUrl)) {27 ????????????????????redirctUrl = request.getRequestURI();28 ????????????????}29 ????????????????controller.setAttr("message", msg);30 ????????????????controller.setAttr("stack", exception);31 ????????????????controller.setAttr("redirctUrl", redirctUrl);32 ????????????????controller.render("/WEB-INF/views/error/500.jsp");33 ????????????}34 ????????} finally {35 36 ????????}37 ????????return;38 39 ????}40 41 }