过滤器不初始化EntityManager [英] Filter do not initialize EntityManager
问题描述
我尝试在View模式中使用Open Session，但每次我尝试在我的 ManagedBean EntityManager > entityManager 来 NULL 这是我正在做的事情：
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？
UPDATE
在每个地方搜索有关JPA，Hibernate和EJB之后，最后我找到了一个关于它的好教程（按照这个顺序来理解正在做什么，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
这仅适用于您的 PlayerBean 也是请求作用域。如果它是视图作用域，则忽略任何手动创建的请求范围属性，而不是仅仅因为不允许此构造而注入。您只能注入与接受者相同或更广范围的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.
我根据您的问题历史知道您正在使用Glassfish 3.为什么不知道你刚刚使用EJB吗？这样容器就会担心事务本身，你根本不需要这样的过滤器。您可以通过 EntityManager noreferrer> @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.
这很简单。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无需进一步配置）
然后，您可以在JSF托管bean中使用它：
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！
