如何防止一个ID在不同的机器上同进登录?

Question

创建一个类，实现HttpSessionBindingListener，然后在
public void valueUnbound(HttpSessionBindingEvent event) {
//     在此方法里做用户退出时的处理
    }把此类实例化，然后放在用户的session里。用户退出时会调用valueUnbound

Answer 1

我的类如下：import java.io.Serializable;
import javax.servlet.http.*;/**
 * Database implementation of the Authorization interface.
 */
public final class Authorization implements Serializable, HttpSessionBindingListener {    private static final long serialVersionUID = 01L;    private int userID;    /**
     * Constucts a new DbAuthorization with the specified userID.
     *
     * @param userID the userID to create an authorization token with.
     */
    protected Authorization(int userID) {
        this.userID = userID;
    }    //FROM THE AUTHORIZATION INTERFACE//    public int getUserID() {
        return userID;
    }    public boolean isAnonymous() {
        return userID == -1;
    }    /*static int loginUserCount = 0;    public static int getLoginUserCount() {
        return loginUserCount;
    }
*/
    public void valueBound(HttpSessionBindingEvent event) {
       // loginUserCount++;
    }    public void valueUnbound(HttpSessionBindingEvent event) {
       // loginUserCount--;
        UserLoginManage.remove(userID);
        //System.out.println("userID:"+userID);
        LogManage.addLog(new Log(userID,UserManage.getUser(userID).getUnitID(),"",null,
                    "login.jsp?logout=true","离开","UserLogin",0,"","session过期，自动离开",0));
    }
}