Question

我正在开发一个java servlet。我正在使用glassfish服务器4.

最终用户通过网址参数向我发送信息，如下所示：

http://myIP:8080/TestProject/TestServlet?param1=test1&param2=test2&param3=test3

我从param1，param2和param3获取值，我想在我的数据库中写入它们。如果我在我的数据库中写入信息时遇到SQL异常，我想抛出“500内部服务器错误”让他们知道我有一些技术问题并重新发送他们的请求。我想知道是否有默认方式来执行此操作，设置一些状态，显示文本......？

以下是代码：

@WebServlet(urlPatterns = {"/TestServlet"}, initParams = {
    @WebInitParam(name = "param1", value = ""),
    @WebInitParam(name = "param2", value = ""),
    @WebInitParam(name = "param3", value = "")})
public class TestServlet extends HttpServlet {

    String param1;
    String param2;
    String param3;
    boolean dbOK;

    /**
     * Processes requests for both HTTP
     * <code>GET</code> and
     * <code>POST</code> methods.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    protected void processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        //Get parametars from the request
        param1 = request.getParameter("param1");
        param2 = request.getParameter("param2");
        param3 = request.getParameter("param3");
        //Input in db
        dbOK = Database.saveParams(param1,param2,param3);

        //Print response
        response.setContentType("text/html;charset=UTF-8");
        PrintWriter out = response.getWriter();
        try {
            /* TODO output your page here. You may use following sample code. */
            out.println("<!DOCTYPE html>");
            out.println("<html>");
            out.println("<head>");
            out.println("<title>T-Mobile Interface</title>");            
            out.println("</head>");
            out.println("<body>");
            out.println("<h1> dbOK=" + dbOK + "</h1>");
            out.println("</body>");
            out.println("</html>");
        } finally {            
            out.close();
        }
    }

    // <editor-fold defaultstate="collapsed" desc="HttpServlet methods. Click on the + sign on the left to edit the code.">
    /**
     * Handles the HTTP
     * <code>GET</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Handles the HTTP
     * <code>POST</code> method.
     *
     * @param request servlet request
     * @param response servlet response
     * @throws ServletException if a servlet-specific error occurs
     * @throws IOException if an I/O error occurs
     */
    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    /**
     * Returns a short description of the servlet.
     *
     * @return a String containing servlet description
     */
    @Override
    public String getServletInfo() {
        return "Short description";
    }// </editor-fold>
}

Answer 1

您应该可以使用response.sendError(int)

执行此操作

编辑：意味着该参数是您要发送的错误代码，因此在您的情况下为500。

Answer 2

AFAIK ServletException也被翻译为错误500。但通常您需要在响应上设置状态并将其重定向（调度）到JSP，其中包含您的自定义错误页面。您可以设置请求属性以自定义JSP内容。

req.setAttribute("MY_ERROR", "Database request failed");
resp.setStatus(HttpServletResponse.SC_ERROR);
ServletContext context = getServletContext();
RequestDispatcher dispatcher = context.getRequestDispatcher("/error.jsp");
dispatcher.forward(req, resp);

Java servlet故意抛出http 500错误

2 个答案: