Question

我这几个小时一直在努力。

我正试图发一条推文（twit）

上下文：

我在这个webapp上有多个servlet（应该是Twitter的极简主义版本）。

web.xml的片段如下所示：

<welcome-file-list>
    <welcome-file>login.jsp</welcome-file>
</welcome-file-list>
<servlet>
    <servlet-name>membershipServlet</servlet-name>
    <servlet-class>controller.membershipServlet</servlet-class>  
</servlet>   
<servlet>
    <servlet-name>twitServlet</servlet-name>    
    <servlet-class>controller.twitServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>membershipServlet</servlet-name>
    <url-pattern>/membership</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>twitServlet</servlet-name>
    <url-pattern>/twitpost</url-pattern>
</servlet-mapping>
<session-config>
    <session-timeout>
        30
    </session-timeout>
</session-config>
</web-app>

作业说要做一个推文的doPost。

应该控制它的servlet片段如下所示：

24 @WebServlet(name = "twitServlet", urlPatterns = {"/twitsubmit"}) 
25 public class twitServlet extends membershipServlet { 
26  
27      /** 
28      * Handles the HTTP <code>POST</code> method. 
29      * 
30      * @param request servlet request 
31      * @param response servlet response 
32      * @throws ServletException if a servlet-specific error occurs 
33      * @throws IOException if an I/O error occurs 
34      */ 
35     @Override 
36     protected void doPost(HttpServletRequest request, HttpServletResponse 
       response) 
37             throws ServletException, IOException, NullPointerException { 
38  
39         String action = request.getParameter("action"); 
40         String url = "/home.jsp";  // Main start page // 
41         HttpSession session = request.getSession(); 
42  
43         if (action.equals("twitpost")) { 
44             //get paramerterws 
45             String twittext = request.getParameter("twitSubmit"); 
46
47             //store in the twit object   
48             Twit mytwit = new Twit(); 
49             mytwit.setTwitText(twittext); 
50  
51  
52             session.setAttribute("enteredTwit", twittext); 
53             TwitDB.twit(twittext); 
54             request.setAttribute("twit", twittext); 
55             url = "/home.jsp"; 
56             session.setAttribute("twit", mytwit); 
57             session.removeAttribute("enteredTwit"); 
58         } 
59  
60  
61         getServletContext() 
62                 .getRequestDispatcher(url) 
63                 .forward(request, response); 
64     }

尝试向数据库输入推文最终会出现500错误：

HTTP Status 500 – Internal Server Error

Type Exception Report

Description The server encountered an unexpected condition that prevented it 
from fulfilling the request.

Exception

java.lang.NullPointerException
controller.twitServlet.doPost(twitServlet.java:44)
javax.servlet.http.HttpServlet.service(HttpServlet.java:660)
javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)

如果有人能提供帮助，我将永远感激不尽。我一直在努力比这更长时间。

谢谢。

Java Servlet提供HTTP状态500

0 个答案: