02. 将异常堆栈信息，返回给前台，便于排查问题.

Posted 2020-10-02

需求：

1. 我们在排查线上可复现问题时，如果出现异常可能还需要登陆服务器去翻日志.

解决：

1. java Code

   @RequestMapping("UserController/register")
   @ResponseBody
   public Map<Object, Object> register(String email, String userName, String password) {
   	HashMap<Object, Object> result = MethodReturn.returnMapDefault();
   	try {
   		// result = userService.register(email, userName, password);
   		int a = 1 / 0;
   	} catch (Exception e) {
   		e.printStackTrace();
   		/*********************************未便于展示，暂为封装******************************************/
   		Map<Object, String> exceptionStatckTrace = new LinkedHashMap<>();
   		exceptionStatckTrace.put(0, e.getClass().getName() + ": " + e.getMessage());
   		StackTraceElement[] stackTrace = e.getStackTrace();
   		int length = stackTrace.length;
   		StackTraceElement obj = null;
   		StringBuilder fileName = null;
   		for (int i = 0; i < length; i++) {
   			obj = stackTrace[i];
   			fileName = new StringBuilder();
   			if (obj.getFileName() != null) {
   				fileName.append(obj.getFileName());
   				if (obj.getLineNumber() != -1) {
   					fileName.append(": " + obj.getLineNumber());
   				}
   			} else {
   				fileName.append("Unknown Source");
   			}
   			exceptionStatckTrace.put(i + 1, obj.getClassName() + "." + obj.getMethodName() + "(" + fileName.toString() + ")");
   		}
   		/***************************************************************************/
   		result.put("exception", exceptionStatckTrace);
   	} finally {
   		return result;
   	}
   }

2. 效果
   

2017年8月13日 15:19:18

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