过滤器不初始化 EntityManager [英] Filter do not initialize EntityManager
问题描述
我尝试在视图模式中使用打开会话，但是每次我尝试在我的 ManagedBean 中捕获 EntityManager 时，entityManager 都会<代码>NULL 这是我的做法:
I trying to use the Open Session in View pattern, but everytime I try to catch the EntityManager in my ManagedBean the entityManager come NULL here is how I'm doing:
package filters;
// imports..
public class JPAFilter implements Filter {
private EntityManagerFactory factory;
@Override
public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException {
EntityManager entityManager = this.factory.createEntityManager();
request.setAttribute("entityManager", entityManager);
entityManager.getTransaction().begin();
chain.doFilter(request, response);
try {
entityManager.getTransaction().commit();
} catch (Exception e) {
entityManager.getTransaction().rollback();
throw new ServletException(e);
} finally {
entityManager.close();
}
}
@Override
public void init(FilterConfig filterConfig) throws ServletException {
this.factory = Persistence.createEntityManagerFactory("copadomundo");
}
@Override
public void destroy() {
this.factory.close();
}
}
这是我的 ManagedBean:
And this is my ManagedBean:
package managedbeans;
// imports ..
@ManagedBean
public class PlayerBean {
@ManagedProperty(value = "#{entityManager}")
private EntityManager entityManager;
private Player player = new Player();
private Long teamID;
private List<Player> players = new ArrayList<Player>();
public void add() {
TeamRepository selecaoRepository = new TeamRepository(this.entityManager);
Team selecao = selecaoRepository.search(this.teamID);
this.player.setTeam(selecao);
PlayerRepository playerRepository = new PlayerRepository(this.entityManager);
playerRepository.adiciona(this.player);
this.player = new Player();
this.players = null;
}
public void remove(Player player) {
PlayerRepository repository = new PlayerRepository(this.entityManager);
repository.remove(player);
this.players = null;
}
// GETTERS AND SETTERS
public List<Player> getPlayeres() {
if (this.players == null) {
PlayerRepository repository = new PlayerRepository(
this.entityManager);
this.players = repository.getPlayeres();
}
return this.players;
}
public EntityManager getEntityManager() {
return entityManager;
}
public void setEntityManager(EntityManager entityManager) {
this.entityManager = entityManager;
}
public Player getPlayer() {
return player;
}
public void setPlayer(Player player) {
this.player = player;
}
public Long getTeamID() {
return teamID;
}
public void setTeamID(Long teamID) {
this.teamID = teamID;
}
public void setPlayeres(List<Player> players) {
this.players = players;
}
}
这是我的 web.xml:
And this is my web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
id="WebApp_ID" version="3.0">
<display-name>WorldCup</display-name>
<welcome-file-list>
<welcome-file>index.xhtml</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Faces Servlet</servlet-name>
<servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Faces Servlet</servlet-name>
<url-pattern>*.xhtml</url-pattern>
</servlet-mapping>
<filter>
<filter-name>LoginFilter</filter-name>
<filter-class>jpa.LoginFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>LoginFilter</filter-name>
<servlet-name>Faces Servlet</servlet-name>
</filter-mapping>
<filter>
<filter-name>JPAFilter</filter-name>
<filter-class>jpa.JPAFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>JPAFilter</filter-name>
<servlet-name>Faces Servlet</servlet-name>
</filter-mapping>
<error-page>
<exception-type>java.lang.Exception</exception-type>
<location>/error.xhtml</location>
</error-page>
</web-app>
知道为什么会这样吗?
更新在到处搜索关于 JPA、Hibernate 和 EJB 之后，我终于找到了一个很好的教程(按照这个顺序来了解正在做什么，好吗?):
UPDATE After searching in every place about JPA, Hibernate and EJB, finally I found a good tutorial about it (follow this order to understand what is been doing, okay ?):
为 Eclipse 和 Oracle Glassfish 3.1 安装和配置 MySQL
推荐答案
这仅在您的 PlayerBean 也是请求范围时才有效.如果它是视图范围的，那么任何手动创建的请求范围的属性都将被忽略并且不会仅仅因为不允许这种构造而注入.您只能注入与接受器相同或更广泛范围的 JSF 托管属性.
That will only work if your PlayerBean is also request scoped. If it is view scoped, then any manually created request scoped attributes are ignored and not injected simply because this construct is not allowed. You can only inject a JSF managed property of the same or broader scope than the acceptor.
根据您的问题历史，我知道您使用的是 Glassfish 3.为什么不直接使用 EJB?这样容器将担心事务本身，您根本不需要这样的过滤器.您可以通过 EntityManager>@PersistenceContext.
I know based on your question history that you're using Glassfish 3. Why don't you just use an EJB? This way the container will worry about transactions itself and you don't need to have such a filter at all. You can inject the EntityManager by @PersistenceContext.
很简单.只需创建以下 EJB 类:
It's pretty simple. Just create the following EJB class:
@Stateless
public class PlayerService {
@PersistenceContext
private EntityManager em;
public Player find(Long id) {
return em.find(Player.class, id);
}
public List<Player> list() {
return em.createQuery("SELECT p FROM Player p", Player.class).getResultList();
}
public void create(Player player) {
em.persist(player);
}
public void update(Player entity) {
em.merge(player);
}
public void delete(Player player) {
em.remove(em.contains(player) ? player : em.merge(player));
}
// ...
}
(Glassfish 3 上无需进一步配置)
然后您可以在 JSF 托管 bean 中按如下方式使用它:
You can then use it as follows in your JSF managed bean:
@ManagedBean
@ViewScoped
public class PlayerBean {
private List<Player> players;
@EJB
private PlayerService playerService;
@PostConstruct
public void init() {
players = playerService.list();
}
// ...
}
这篇关于过滤器不初始化 EntityManager的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
