spring - java有什么方法可以捕获程序异常的？然后中断程序，发送信息给前端-Fragen und Antworten zum chinesischen PHP-Netzwerk

我现在有这个问题：

我在这里想获取整数的，但是文件中出现了小数，那么就有异常了这里，如何抓获这个异常呢？try catch？

cell.setCellType(Cell.CELL_TYPE_STRING); result = cell.getStringCellValue();

java.lang.NumberFormatException: For input string: "103.12"

当出现这个问题的时候，程序是中断了，但是只是在控制台输出这个信息。如果用什么方法捕获这些错误，然后发送信息给前端呢？

现在出现这个错误，前端就显示卡死状态。除非关闭页面。

控制台也停了：

只是tomcat自己获取错误：

try { cell.setCellType(Cell.CELL_TYPE_STRING); result = cell.getStringCellValue(); } catch (NumberFormatException e) { e.printStackTrace(); }

抓获异常，怎么抓呢？在哪里返回异常信息给前端呢？

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请问呢，如何在后端java中获取response呢？我用的是spring和springmvc。更具体说：怎么在catch上面获得response，并且输入信息呢？

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try { cell.setCellType(Cell.CELL_TYPE_STRING); result = cell.getStringCellValue(); } catch (NumberFormatException e) { System.out.println("转换发生如下错误："+e.getMessage()); try { String message = e.getMessage(); response.getWriter().write(message); } catch (IOException e1) { e1.printStackTrace(); } e.printStackTrace(); }

我都这样写了，还是在控制台什么没输出。System.out.println("转换发生如下错误："+e.getMessage());这句代码没用啊！！！停住在这里....

public static boolean test() { try { int i = 10 / 0; // 抛出 Exception，后续处理被拒绝 System.out.println("i vaule is : " + i); return true; // Exception 已经抛出，没有获得被执行的机会 } catch (Exception e) { System.out.println(" -- Exception --"); return catchMethod(); // Exception 抛出，获得了调用方法的机会，但方法值在 finally 执行完后才返回 }finally{ finallyMethod(); // Exception 抛出，finally 代码块将在 catch 执行 return 之前被执行 } }

@ControllerAdvice public class GlobalExceptionHandlingControllerAdvice { protected Logger logger; public GlobalExceptionHandlingControllerAdvice() { logger = LoggerFactory.getLogger(getClass()); } /* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - */ /* . . . . . . . . . . . . . EXCEPTION HANDLERS . . . . . . . . . . . . . . */ /* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - */ /** * Convert a predefined exception to an HTTP Status code */ @ResponseStatus(value = HttpStatus.CONFLICT, reason = "Data integrity violation") // 409 @ExceptionHandler(DataIntegrityViolationException.class) public void conflict() { logger.error("Request raised a DataIntegrityViolationException"); // Nothing to do } /** * Convert a predefined exception to an HTTP Status code and specify the * name of a specific view that will be used to display the error. * * @return Exception view. */ @ExceptionHandler({ SQLException.class, DataAccessException.class }) public String databaseError(Exception exception) { // Nothing to do. Return value 'databaseError' used as logical view name // of an error page, passed to view-resolver(s) in usual way. logger.error("Request raised " + exception.getClass().getSimpleName()); return "databaseError"; } /** * Demonstrates how to take total control - setup a model, add useful * information and return the "support" view name. This method explicitly * creates and returns * * @param req * Current HTTP request. * @param exception * The exception thrown - always {@link SupportInfoException}. * @return The model and view used by the DispatcherServlet to generate * output. * @throws Exception */ @ExceptionHandler(SupportInfoException.class) public ModelAndView handleError(HttpServletRequest req, Exception exception) throws Exception { // Rethrow annotated exceptions or they will be processed here instead. if (AnnotationUtils.findAnnotation(exception.getClass(), ResponseStatus.class) != null) throw exception; logger.error("Request: " + req.getRequestURI() + " raised " + exception); ModelAndView mav = new ModelAndView(); mav.addObject("exception", exception); mav.addObject("url", req.getRequestURL()); mav.addObject("timestamp", new Date().toString()); mav.addObject("status", 500); mav.setViewName("support"); return mav; } }

Antworte allen (3)

迷茫2017-04-18 09:42:43 3 Etage

1 你catch这个exception
2 在你的catch里你返回一个消息给response.getOutputStream或者response.getWriter
3 你前端的js（如果你用的是ajax）读取步骤2里的输出，提给给前端；如果你是jsp，那么就在jsp里out.print(xxx);

Wie+0

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