Question

我想从html页面向servlet发送请求，然后在java servlet中创建一个列表，然后我想将此列表返回到同一个html页面。

Answer 1

问题是您永远不会初始化HttpSession s变量：

@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);
    }
}

Make-it-work 解决方案：设置s

的值

s = request.getSession();

真实世界解决方案：

删除 Servlet中的所有字段， NEVER 尝试处理servlet中的状态，除非它们是容器管理的资源，如EJB。
将HttpSession s的范围更改为每种方法的本地范围。另外，将其名称从s更改为session或更有用的内容。
将HTML代码移动到处理视图详细信息（如JSP）的组件，然后执行转发到视图。

所以，你的代码看起来像这样：

@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：

Answer 2

您需要从request获取会话，您不能简单地期望通过HttpSession s ;获取会话 - s为null。

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

Answer 3

您需要提供异常的堆栈跟踪。但它看起来可能在两个地方：

s.setAttribute（＆＃34;人＆＃34;，人）;其中s从未设置;
out.println（＆＃34;＆＃34; + p.getKey（）+＆＃34;＆＃34;）;属性＆＃34; person＆＃34;永远不会被设定;

Answer 4

未定义var s的值。它为null，因此使用s.getAttribute("str")会抛出异常。

HttpSession s ;

s = request.getSession(false);

有没有办法将列表从servlet返回到html页面

4 个答案: