有沒有辦法從servlet返回一個列表到html頁面

Question

我想從html頁面發送請求到servlet然後在java servlet中創建一個列表然後我想把這個列表返回到同一個html頁面。

Answer 1

的問題是，你永遠不會初始化HttpSession s變量：

@WebServlet(urlPatterns = {"/ShowPersonServlet"}) 
public class ShowPersonServlet extends HttpServlet { 

    HttpSession s; //null by default 
    //... 

    protected void processRequest(...) { 
     //... 
     //since never initialized, s is null 
     user.add((Person) s.getAttribute("person")); 
    } 

    @Override 
    protected void doGet(...) { 
     //... 
     //since never initialized, s is null 
     s.setAttribute("person",person); 
    } 
} 

化妝它工作解決方案：設置s

s = request.getSession(); 

現實世界解決方案的價值：

• 刪除 Servlet中的所有字段，從不嘗試處理servlet中的狀態，除非它們是由容器（如EJB）管理的資源。
• 將您的HttpSession s的範圍更改爲每個方法的本地範圍。另外，將其名稱從s更改爲session或更有用的東西。
• 將您的HTML代碼移動到處理查看詳細信息（如JSP）的組件，然後執行轉發視圖。

所以，你的代碼應該是這樣的：

@WebServlet(urlPatterns = {"/ShowPersonServlet"}) 
public class ShowPersonServlet extends HttpServlet { 
    /* 
    HttpSession s ; //moved as local variable 
    Person person = new Person(); //moved as local variable 
    private List<Person> user = new ArrayList<Person>(); //not sure where you want to store this 
    */ 
    protected void processRequest(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     response.setContentType("text/html;charset=UTF-8"); 
     HttpSession session = request.getSession(); 
     List<Person> personList = (List<Person>) session.getAttribute("personList"); 
     if (personList == null) { 
      personList = new ArrayList<>(); 
      session.setAttribute("personList", personList); 
     } 
     personList.add((Person) session.getAttribute("person")); 
     /* 
     try (PrintWriter out = response.getWriter()) { 
      //removed to shorten this answer 
     } 
     */ 
     request.getRequestDispatcher("/showPerson.jsp").forward(request, response); 
    } 

    @Override 
    protected void doGet(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     Person person = new Person(); 
     person.setKey(request.getParameter("txt_Key")); 
     person.setFirstName(request.getParameter("txt_firstName")); 
     person.setLastName(request.getParameter("txt_lastName")); 
     processRequest(request, response); 
     HttpSession session = request.getSession(); 
     session.setAttribute("person",person); 
    } 

    @Override 
    protected void doPost(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     processRequest(request, response); 
    } 
} 

更多信息：

• How do servlets work? Instantiation, sessions, shared variables and multithreading
• StackOverflow Servlets wiki

Answer 2

你需要得到來自會話request，你c annot只是期望得到它與HttpSession s ; - 有s是null。

// Something like this (in processRequest), although I 
// would prefer a local session variable. 
s = request.getSession(); 

Answer 3

您需要提供異常的堆棧跟蹤。但它看起來可能在兩個地方：

1. s.setAttribute（「person」，person）; s從未設置;
2. out.println（「」+ p.getKey（）+「」）;其中屬性「人」從未設置;

祝你好運。

Answer 4

var s的值未定義。它是空的，所以使用s.getAttribute("str")將會拋出異常。

HttpSession s ;

s = request.getSession(false);