Java Servlet nullPointerException登錄註銷會話

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嗨我是Servlets新手可能很容易的問題。在我將數據傳遞到表單並單擊提交按鈕後，我得到java.lang.NullPointerExceptionLogOut.doPost(LogOut.java:45)任何想法如何通過此異常？Java Servlet nullPointerException登錄註銷會話

代碼的doGet：

protected void doGet(HttpServletRequest request, HttpServletResponse response) { 

    PrintWriter out = response.getWriter(); 
    response.setContentType("text/html"); 
    HttpSession session = request.getSession(false); 

    out.println("<form method='post' action='LogOut'>" 
      + "usr <input type='text' id='username'/>" 
      + "pass <input type='password'id='password'/>" 
      + "<input type='submit' value='Login'></form>"); 

    //If the username and password is correct 
    if (session != null) { 
     String username = (String) session.getAttribute("username"); 
     out.println("Hi " + username + "sessioon is started with id " + session.getId()); 
     out.println("<a id='logout' value='Logout' />"); 
    } 
}

代碼的doPost

protected void doPost(HttpServletRequest request, HttpServletResponse response) { 

    PrintWriter out = response.getWriter(); 
    String username = request.getParameter("username"); 
    String password = request.getParameter("password"); 
    response.setContentType("text/html"); 

    if (username.equals("usr") && password.equals("pass")) { //line 45 
     HttpSession sess = request.getSession(); 
     sess.setAttribute("username", username); 
    } else { 
     out.println("Wrong input data"); 
    } 
}

來源

2014-09-28 user3024888

哪條線是45號線？ – deKajoo 2014-09-28 14:35:01

in the doPost method theres // line 45 – user3024888 2014-09-28 14:37:46

檢查request.getParameter（「username」）;'或'request.getParameter（「password」）;'不返回null – deKajoo 2014-09-28 14:39:21

我建議你使用try-catch和捕獲異常，並能編寫用於當異常被捕獲，處理這個問題。像給用戶一個錯誤信息。如果總是出現錯誤消息，那麼代碼中可能存在邏輯錯誤。但讓我困惑的是，第45行沒有代碼。看起來好像你留下了空間。

來源

2014-09-28 14:37:20 rert588

Java Servlet nullPointerException登錄註銷會話

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