非Controller异常处理办法

Posted 2021-03-03 nyf828

最近使用shiro作为权限认证框架,每次通过校验token判断用户是否登录,但是发现一个问题, 自定义的JwtFilter无法通过@ControllerAdvice进行异常捕获(还没到controller呢),但是需要返回统一格式数据

@Slf4j
public class JwtFilter extends BasicHttpAuthenticationFilter {

	/**
	 * 执行登录认证
	 *
	 * @param request
	 * @param response
	 * @param mappedValue
	 * @return
	 */
	@Override
	protected boolean isAccessAllowed(ServletRequest request, ServletResponse response, Object mappedValue) {
		try {
			log.info("Filter=====isAccessAllowed========");
			executeLogin(request, response);
		} catch (Exception e) {
			HttpServletResponse httpServletResponse = (HttpServletResponse) response;
			httpServletResponse.setCharacterEncoding("utf-8");
			try {
				JSONObject jsonObject = new JSONObject();
				jsonObject.put("code", HttpStatus.HTTP_UNAUTHORIZED);
				jsonObject.put("msg", "Token失效 请重新登陆");
				jsonObject.put("data","");
				PrintWriter outputStream = response.getWriter();
				outputStream.write(jsonObject.toJSONString());
				outputStream.close();


			} catch (IOException ex) {
				log.error("错误"+ex);
			}

		}
		return true;
	}
}

为了实现返回统一数据就想了如下办法,直接在filter中响应数据到前端

HttpServletResponse httpServletResponse = (HttpServletResponse) response;
	httpServletResponse.setCharacterEncoding("utf-8");
	try {
		JSONObject jsonObject = new JSONObject();
		jsonObject.put("code", HttpStatus.HTTP_UNAUTHORIZED);
		jsonObject.put("msg", "Token失效 请重新登陆");
		jsonObject.put("data","");
		PrintWriter outputStream = response.getWriter();
		outputStream.write(jsonObject.toJSONString());
		outputStream.close();
	  } catch (IOException ex) {
		log.error("错误"+ex);
		}

如果有更好的办法再来更新这篇博客