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

Question

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？

============================================================

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？

============================================================

    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....

Answer 1

1 Vous interceptez cette exception
2 Dans votre catch, vous renvoyez un message à réponse.getOutputStream ou réponse.getWriter
3 Votre js front-end (si vous utilisez ajax) lit l'étape 2 La sortie est fourni au front-end ; si vous êtes jsp, alors out.print(xxx);

in jsp

Answer 2

Utilisez évidemment Java spécifiquement pour intercepter les exceptions try catch

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
    }
}

Answer 3

Essayez ça ?

@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;
    }
}

Code source