相关疑难解决方法(0)

我有一场包含以下内容的战争:

META-INF/MANIFEST.MF
WEB-INF/classes/META-INF/persistence.xml
WEB-INF/classes/com/test/service/TestServlet.class
WEB-INF/classes/com/test/service/TestEntity.class
WEB-INF/classes/jndi.properties
WEB-INF/classes/postgresql-ds.xml
WEB-INF/jboss-web.xml
WEB-INF/web.xml
index.jsp

persistence.xml中:

<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd" version="1.0">
    <persistence-unit name="test">
        <provider>org.hibernate.ejb.HibernatePersistence</provider>
        <jta-data-source>java:/TestDS</jta-data-source>

        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQLDialect"/>
            <property name="hibernate.hbm2ddl.auto" value="update" />
            <property name="hibernate.show_sql" value="true" />
        </properties>
    </persistence-unit>
</persistence>

web.xml中:

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
  <display-name>Test Web Application</display-name>

    <context-param>
        <param-name>resteasy.scan</param-name>
        <param-value>true</param-value>
    </context-param>
    <listener>
        <listener-class>org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap</listener-class>
    </listener>
    <servlet>
        <servlet-name>Resteasy</servlet-name>
        <servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Resteasy</servlet-name>
        <url-pattern>/service/*</url-pattern>
    </servlet-mapping>

    <resource-ref>
        <res-ref-name>TestDS</res-ref-name>
        <res-type>javax.sql.DataSource</res-type>
        <res-auth>Container</res-auth>
        <res-sharing-scope>Shareable</res-sharing-scope>
    </resource-ref>
</web-app>

我的TestServlet类如下:

package com.test.service; …

我正在尝试使用 CDI 将 EntityManager 注入到我的应用程序中，但在尝试使用它时 EntityManager 为空。

这是我的代码，我遵循了几个有关如何注入 EntityManager 的教程，并且我使用与这些教程中相同的代码。

@Qualifier
@Retention(RetentionPolicy.RUNTIME)
@Target({ElementType.FIELD})
public @interface DevDatabase {

}


@Singleton
public class JPAResourceProducer {

    @Produces
    @PersistenceContext(unitName = "DevPU")
    @DevDatabase
    private EntityManager em;
}

持久化.xml

<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
  <persistence-unit name="DevPU" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <class>entity.MyEntity</class>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties>
      <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/MyDB"/>
      <property name="javax.persistence.jdbc.user" value="appuser"/>
      <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
      <property name="javax.persistence.jdbc.password" value="apppassword"/>
    </properties>
  </persistence-unit>
</persistence>

这就是我在 DAO 中使用它的方式

public abstract class GenericDAO<T> {

    @DevDatabase
    @Inject
    private EntityManager em;
    private final Class<T> entityClass;

    public GenericDAO(Class<T> …