我在网络应用程序中遇到困难，我发布了代码并定义了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，管理员，用户）。在此，我在标记为...的行上收到错误消息。
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，其中包含有关人员的记录
@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 ...这里临时的Pusername，Pname，Pusertype和Pusername用于设置会话的属性。根据此页面的指示，JSP的定向取决于用户类型...（如果用户进入用户主目录，如果经理进入主目录并继续前进）...我知道为什么我会收到此错误，但我不知道避免该错误。我做了研究，对我没有任何帮助...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);
}
如果您希望我添加更多信息，我可以做到。我的问题是，我该如何避免这种情况并使之起作用。
最佳答案
tmp是类型Person的对象。大概，.getType()为您提供String type;类的Person属性。因此，您实际上是在这里设置String对象：
session.setAttribute("usertype", tmp.getType());
因此，以下行结果为
ClassCastException：Person user = (Person) session.getAttribute("usertype");
您需要将返回值强制转换为
String。String userType = (String) session.getAttribute("usertype");
更好的是，您可以在
Person中设置整个session对象。HttpSession session = request.getSession();
session.setAttribute("person", tmp);
然后，您可以将其属性检索为：
Person user = (Person) session.getAttribute("person");
String personType = user.getType();
......................
关于java - 过滤器中的ClassCastException？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/18309148/