Java Web应用程序会话变量
当前用户登录后,如何将其信息存储在会话变量中?这些数据需要在登录后通过web应用程序访问 这就是我到目前为止所做的: User.javaJava Web应用程序会话变量,java,session,Java,Session,当前用户登录后,如何将其信息存储在会话变量中?这些数据需要在登录后通过web应用程序访问 这就是我到目前为止所做的: User.java @Named @Table @Entity @SessionScoped public class User implements Serializable { private static final long serialVersionUID = 1L; @Id @GeneratedValue(strategy=GenerationTy
@Named
@Table
@Entity
@SessionScoped
public class User implements Serializable {
private static final long serialVersionUID = 1L;
@Id @GeneratedValue(strategy=GenerationType.AUTO)
@Column(name = "id")
private long id;
@Column(name = "firstname")
private String firstname;
@Column(name = "lastname")
private String lastname;
@Column(name="username")
private String username;
@Column(name = "password")
private String password;
public User() {}
...
@Named
@RequestScoped
public class UserData implements Serializable {
private static final long serialVersionUID = 1L;
SessionFactory sessionFactory = HibernateUtil.getSessionFactory();
Session session = null;
Transaction tx = null;
List<User> users;
String username;
String password;
public String getUsername() {
return username;
}
public String getPassword() {
return password;
}
public void setUsername(String username) {
this.username = username;
}
public void setPassword(String password) {
this.password = password;
}
public UserData() {
try {
this.session = sessionFactory.openSession();
}
catch(Exception e) {
e.printStackTrace();
}
}
public void login() {
System.out.println("Authorize: Username: " + this.username + " Password: " + this.password);
Query query = session.createQuery("from User U where U.username = :username and U.password = :password");
query.setParameter("username", this.username);
query.setParameter("password", this.password);
System.out.println(query.list());
}
UserData.java
@Named
@Table
@Entity
@SessionScoped
public class User implements Serializable {
private static final long serialVersionUID = 1L;
@Id @GeneratedValue(strategy=GenerationType.AUTO)
@Column(name = "id")
private long id;
@Column(name = "firstname")
private String firstname;
@Column(name = "lastname")
private String lastname;
@Column(name="username")
private String username;
@Column(name = "password")
private String password;
public User() {}
...
@Named
@RequestScoped
public class UserData implements Serializable {
private static final long serialVersionUID = 1L;
SessionFactory sessionFactory = HibernateUtil.getSessionFactory();
Session session = null;
Transaction tx = null;
List<User> users;
String username;
String password;
public String getUsername() {
return username;
}
public String getPassword() {
return password;
}
public void setUsername(String username) {
this.username = username;
}
public void setPassword(String password) {
this.password = password;
}
public UserData() {
try {
this.session = sessionFactory.openSession();
}
catch(Exception e) {
e.printStackTrace();
}
}
public void login() {
System.out.println("Authorize: Username: " + this.username + " Password: " + this.password);
Query query = session.createQuery("from User U where U.username = :username and U.password = :password");
query.setParameter("username", this.username);
query.setParameter("password", this.password);
System.out.println(query.list());
}
@Named
@请求范围
公共类UserData实现可序列化{
私有静态最终长serialVersionUID=1L;
SessionFactory SessionFactory=HibernateUtil.getSessionFactory();
会话=空;
事务tx=null;
列出用户名单;
字符串用户名;
字符串密码;
公共字符串getUsername(){
返回用户名;
}
公共字符串getPassword(){
返回密码;
}
public void setUsername(字符串用户名){
this.username=用户名;
}
public void setPassword(字符串密码){
this.password=密码;
}
公共用户数据(){
试一试{
this.session=sessionFactory.openSession();
}
捕获(例外e){
e、 printStackTrace();
}
}
公共无效登录(){
System.out.println(“授权:用户名:+this.Username+”密码:+this.Password”);
Query Query=session.createQuery(“来自用户U,其中U.username=:username和U.password=:password”);
query.setParameter(“用户名”,this.username);
query.setParameter(“密码”,this.password);
System.out.println(query.list());
}
返回了正确的用户对象,但我不知道如何在用户的会话数据中保存用户对象
我正在使用的技术(如果有帮助):
- 冬眠
- MySQL
- JPA
- JSF
- 野蝇
用户
类由@命名的
注释管理,因此您只需将当前用户的实例注入您想要使用的类中,即可访问该用户:
@Inject
private User user;
另一种解决方案是使用如下代码:
FacesContext context = FacesContext.getCurrentInstance();
User user = context.getApplication().evaluateExpressionGet(context, "#{user}", User.class);
您使用的是哪个web框架?@redflar3我没有使用web框架,这是一个普通的javaEE。请您解释一下使用
evaluateExpressionGet(…)
还是使用context.getAttributes().get(“user”)