java - 过滤器中的 ClassCastException？

Question

我在网络应用程序中遇到困难，我发布代码并定义我在代码部分中所做的事情...

这是我的过滤器，用于检查用户类型(管理员、经理、用户)。我在标记的行处收到错误...

    public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {


    HttpServletRequest req = (HttpServletRequest) request;  
    HttpSession session = req.getSession();
    RequestDispatcher rd=null;

    Person user = (Person) session.getAttribute("usertype"); <------ **IM GETTING EXCEPTION HERE!**

    if (user != null && user.getType().equals(UserType.MANAGER.toString())) {

        String nextJSP = "/ManagerHome.jsp";
        rd = request.getRequestDispatcher(nextJSP);
        rd.forward(request, response);
    }

    else if (user != null && user.getType().equals(UserType.ADMIN.toString())) {

        String nextJSP = "/AdminHome.jsp";
        rd = request.getRequestDispatcher(nextJSP);
        rd.forward(request, response);

    }

    else if (user != null && user.getType().equals(UserType.USER.toString())) {

        String nextJSP = "/UserHome.jsp";
        rd = request.getRequestDispatcher(nextJSP);
        rd.forward(request, response);
    }   
    else {

        String nextJSP = "/Login.jsp";
        rd = request.getRequestDispatcher(nextJSP);
        rd.forward(request, response);
    }

    chain.doFilter(request, response);
}

这是我的 Person 类，其中包含人员记录

    @Table(name="\"Person\"")
    public class Person implements Serializable {

/**
 * 
 */
private static final long serialVersionUID = 2532993385565282772L;
@Id
@Column(name="id",nullable=false,updatable=false)
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String username;    
private String name;
private String surname;
private String sskno;
private String address;
private String telno;
private String type;

@OneToMany
private List<Leave> leaves;

public Person() {
}

     getters & setters....

这是我的LoginServlet...这里暂时Pusername、Pname、Pusertype 和Pusername 用于设置 session 属性。根据此页面，JSP 的定向取决于用户类型...(如果用户转到 userhome，如果经理经理回家并继续)...我知道为什么我会收到此错误，但我不知道如何避免它。我做了研究，但没有任何效果...请帮助我，这是我的 Loginservlet

     public class LoginServlet extends HttpServlet {
private static final long serialVersionUID = 1L;

public LoginServlet() {
    super();
}

protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {
    try {
        String pName;
        String pSurname;
        String pUserName;
        String pUserType;
        String query;
        String home="/Login.jsp";
        String username = request.getParameter("username");
        String password = request.getParameter("password");
        RequestDispatcher rd = request.getRequestDispatcher(home);
        mysqlCon con = new mysqlCon();
        //HttpSession session = request.getSession();
        LoginService ls = new LoginService();

        Statement stmt = con.getConnection().createStatement();
        query = "SELECT name, surname, usertype, username FROM employee WHERE username='"
                + username + "' AND password='" + password + "';";
        stmt.executeQuery(query);
        ResultSet rs = stmt.getResultSet();


        if(rs.next()){

        pName = rs.getString(1);
        pSurname = rs.getString(2);
        pUserType = rs.getString(3);
        pUserName = rs.getString(4);


        if (ls.loginCheck(username, password) != false) {
            Person tmp = new Person();

            tmp.setName(pName);
            tmp.setSurname(pSurname);
            tmp.setType(pUserType);
            tmp.setUsername(pUserName); 

            HttpSession session = request.getSession();
            session.setAttribute("name", tmp.getName());
            session.setAttribute("surname", tmp.getSurname());
            session.setAttribute("usertype", tmp.getType());
            session.setAttribute("username", tmp.getUsername());

             if (pUserType.equals(UserType.MANAGER.toString())) {

                    String nextJSP = "home/ManagerHome.jsp";
                    rd = request.getRequestDispatcher(nextJSP);
                    rd.forward(request, response);
                }

                else if (pUserType.equals(UserType.ADMIN.toString())) {

                    String nextJSP = "home/AdminHome.jsp";
                    rd = request.getRequestDispatcher(nextJSP);
                    rd.forward(request, response);

                }

                else if (pUserType.equals(UserType.USER.toString())) {

                    String nextJSP = "home/UserHome.jsp";
                    rd = request.getRequestDispatcher(nextJSP);
                    rd.forward(request, response);
                }   
                else {

                    String nextJSP = "/Login.jsp";
                    rd = request.getRequestDispatcher(nextJSP);
                    rd.forward(request, response);
                }

        }
        }
        else {

            rd.forward(request, response);

        }

如果您想让我添加更多信息，我可以做到。我的问题是我怎样才能避免这种情况并使其发挥作用。

Answer 1

tmp 是 Person 类型的对象。据推测，.getType() 会获取 Person 类的 String type; 属性。因此，您实际上是在此处设置一个 String 对象:

session.setAttribute("usertype", tmp.getType());

因此下面的行结果为 ClassCastException :

Person user = (Person) session.getAttribute("usertype"); 

您需要将返回值转换为 String 。

String userType = (String) session.getAttribute("usertype");

<小时/>

更好的是，您可以在 session 中设置整个 Person 对象。

HttpSession session = request.getSession();
session.setAttribute("person", tmp);

然后您可以检索其属性:

Person user = (Person) session.getAttribute("person");
String personType = user.getType();
......................