调度程序不会抛出异常

Question

我想让调度程序在我想要转发到不存在的资源时抛出异常，这是我

 String page = (String) request.getAttribute("page");  //page to be forwarded form servlet to jsp
    if (page == null) {
        page = request.getParameter("page");//page to be forwarded form jsp to servlet
    }
    RequestDispatcher dispatcher = request.getRequestDispatcher("/WEB-INF/InstitutionPages/" + page + ".jsp");
    try {
        dispatcher.forward(request, response);
    } catch (IOException ex) {
           ex.printStackTrace();
        LogoutServlet.redirectToLoginPage(request, response);
    } catch (javax.servlet.ServletException e) {
           e.printStackTrace();
        Logger.getLogger(RegistrarManagementServlet.class.getName()).log(Level.SEVERE, null, e);
        LogoutServlet.redirectToLoginPage(request, response);
    } catch (java.lang.IllegalArgumentException e) {
        e.printStackTrace();
        LogoutServlet.redirectToLoginPage(request, response);
    }

在页面中的代码，我发送无效的页面名称，但此错误发生在控制台上

SEVERE: PWC6117: File "D:\versions\v30\OnlineQuerySystem_New\build\web\WEB-INF\InstitutionPages\Registerkk.jsp" not found

没有打印任何堆栈跟踪！

Answer 1

您的 servlet 可能如下所示：

public class SimpleServlet extends HttpServlet {

@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
        throws ServletException, IOException {

    // do something at the servlet here

    String page = (String) req.getAttribute("page"); // page to be forwarded
                                                        // form servlet to
                                                        // jsp
    if (page == null) {
        page = req.getParameter("page");// page to be forwarded form jsp to
                                        // servlet
    }

    this.forwardIfExists(req, resp, page);

}

protected void forwardIfExists(HttpServletRequest req,
        HttpServletResponse resp, String page) throws ServletException, IOException {

    File pagePath = new File(this.getServletContext().getRealPath(page));

    if ( pagePath.exists() ) {
        req.getRequestDispatcher( page ).forward(req, resp);
    } else {
        throw new IllegalArgumentException(String.format( "The page %s does not exist", page ));
    }

}

}

另外，不要捕获 servlet 方法抛出的 ServletException 或 IOException，如果它们发生在您的应用程序中，并且您正在发生一些非常糟糕的事情不应该像在代码中那样吞掉这些异常。这些异常应该保持原样，并且容器应该捕获它们。您应该记录它们，而不是尝试打印堆栈跟踪，因为这将在 err 流中打印，并且在生产服务器上不可见。