Question

我正在尝试在Eclipse中配置示例JPA应用程序，并将其部署到TomEE +。数据源是容器管理的。尝试创建EntityManager时，我一直看到以下错误：

持久性提供程序正在尝试使用persistence.xml文件中的属性来解析数据源。必须在openjpa.ConnectionDriverName或javax.persistence.jdbc.driver属性中指定Java数据库连接（JDBC）驱动程序或数据源类名。配置中提供了以下属性：“org.apache.openjpa.jdbc.conf.JDBCConfigurationImpl@414793b4”。

知道这个配置有什么问题吗？

以下是代码。

tomee.xml

<tomee>
  <Resource id="jdbc/MyAppDS" type="DataSource">
    JdbcDriver          com.microsoft.sqlserver.jdbc.SQLServerDriver
    JdbcUrl             jdbc:sqlserver://localhost:1433/db_dev
    UserName            user
    Password            password
    JtaManaged          true
    DefaultAutoCommit   false
  </Resource>
</tomee>

的persistence.xml

<persistence xmlns="http://java.sun.com/xml/ns/persistence"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    version="2.0">

    <persistence-unit name="Simplest" transaction-type="JTA">
        <provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
        <jta-data-source>jdbc/MyAppDS</jta-data-source>
        <class>social.Media</class>
    </persistence-unit>
 </persistence>

Media.java

package social;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "media")
public class Media {

    @Id
    @Column(name = "id")
    private String id;

    private String description;

    private String title;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

}

Controller.java

package social;

import javax.inject.Inject;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/hello")
public class Controller {

    @Inject private Media media;

    @GET
    @Produces(MediaType.TEXT_PLAIN)
    public String sayHello() {

        EntityManagerFactory emf = Persistence.createEntityManagerFactory("Simplest");

        EntityManager em = emf.createEntityManager(); // exception reported on this line
        .
        .
        .

        return media.getDescription();
    }

}

Answer 1

您正在使用JTA管理的Entity Manager Factory，我认为您应该让应用服务器为您执行此操作，而不是手动创建EntityManagerFactory，就像在控制器中一样：

@Path("/hello")
public class Controller {

  @PersistenceContext(unitName="Simplest")
  private EntityManager em;

  @GET
  @Produces(MediaType.TEXT_PLAIN)
  public String sayHello() {
      // get Media from database - replace with your own code
      Media media = em.find(Media.class, "1");
      return media.getDescription();
  }
}

TomEE + OpenJPA：使用容器管理的DataSource创建EntityManager时出错

1 个答案: