@PersistenceContext EntityManager(JPA)

时间:2017-08-12 18:46:25

标签: java database jpa glassfish ejb

我正在使用JPA和EJB创建Java EE应用程序。 @PersistenceContext(unitName =" airline")中的问题,当我对此行进行注释时,它会给出错误null_pointer_exception,并在代码在服务器上运行时收到以下错误。

Severe:   Exception while preparing the app : Could not resolve a persistence unit corresponding to the persistence-context-ref-name [com.airline.service.Test2EJBService/entityManager] in the scope of the module called [Test2]. Please verify your application.

我的班级:

AddPassenger2

@WebServlet(name = "AddPassenger2", urlPatterns = {"/AddPassenger2"})
public class AddPassenger2 extends HttpServlet {


    @EJB
    Test2EJBService bService;


   protected void processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        response.setContentType("text/html;charset=UTF-8");
        try (PrintWriter out = response.getWriter()) {

            out.println("<!DOCTYPE html>");
            out.println("<html>");
            out.println("<head>");
            out.println("<title>Servlet AddPassenger</title>");
            out.println("</head>");
            out.println("<body>");
            out.println("<h4> Hello Every body in Test2 :) <h4>");
            out.println("</body>");
            out.println("</html>");
        }
    }

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);

        Passenger passenger = new Passenger();

        passenger.setFirstName("Loai");
        passenger.setLastName("Amr");

        Calendar cal = Calendar.getInstance();
        cal.set(Calendar.YEAR, 1996);
        cal.set(Calendar.MONTH, 9);
        cal.set(Calendar.DAY_OF_MONTH, 30);
        Date dob = cal.getTime();

        passenger.setDob(dob);
        passenger.setGender(Gender.Male);
        passenger.setFlightClass(FlightClass.Couch);     

        bService.addPassenger(passenger);

        System.out.println("First name : " + passenger.getFirstName());

    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        processRequest(request, response);
    }

    @Override
    public String getServletInfo() {
        return "Short description";
    }// </editor-fold>

}

我的EJB容器:

@Stateless
@LocalBean
public class Test2EJBService {

//    @PersistenceUnit(unitName = "Test2PU")
    @PersistenceContext(unitName = "airline")
    private EntityManager entityManager;

    public void addPassenger(Passenger passenger) {
//        EntityManager createEntityManager;
//        createEntityManager = Persistence.createEntityManagerFactory("Test2PU").createEntityMana‌ger();
        entityManager.persist(passenger);
        System.out.println("Done");
    }
}

和persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<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="Test2PU" transaction-type="JTA">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <jta-data-source>jdbc/airline</jta-data-source>
    <exclude-unlisted-classes>false</exclude-unlisted-classes>
    <properties>
      <property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/>
    </properties>
  </persistence-unit>
</persistence>

enter image description here

任何身体都可以帮助我在相同的错误中度过3天并且无法解决它。

以及我们与数据库的连接: enter image description here

1 个答案:

答案 0 :(得分:0)

persistence-unit-name文件中的persistence.xml仍然显示“Test2PU”而不是“航空公司”。验证。另外,请确保在服务器上使用相同名称(即航空公司)正确配置了连接池和JDBC资源