java - EntityManager 不持久化实体类？

Question

我正在尝试使用 JPA 在 RESTful WebService 的 Java Web 应用程序中实现持久性。该程序通过 EntityManager 运行，持久化实体类，被事务包围(开始和提交)。虽然交易会带来麻烦。当我检查 MySQL 数据库时，会创建必要的表，但其中没有条目。怎么可能？代码有什么问题？

我的服务类:

@Stateless
public class ReportService {
    @PersistenceContext(unitName = "AbcPU") // default type is PersistenceContextType.TRANSACTION
    EntityManager em;

    public void saveDog() {
        System.out.println("BBBBB Start ReportService.saveDog();");
        Doggy doggy = new Doggy();
        doggy.setDogName("Wuffi");
        try {
            System.out.println("BBBBB Persist Success ReportService.saveDog();");
            em.getTransaction().begin();
            em.persist(doggy);
            em.flush();
            em.getTransaction().commit();
        } catch (Exception ex) {
            System.out.println("BBBBB Persist Fail ReportService.saveDog();");
            System.err.println("Error with em.persist(doggy): " + ex.getMessage());
        }
        System.out.println("BBBBB Stop ReportService.saveDog();");
    }
}

我的资源类:

@Path("report")
@Produces(MediaType.APPLICATION_JSON)
public class ReportResource {
    @EJB
    private ReportService rs;

    @GET
    public Response findReports() {
        final List<Report> reports = rs.findAllReports();

        System.out.println("AAAAA Start rs.saveDog();");
        rs.saveDog();
        System.out.println("AAAAA Stop rs.saveDog();");

        return Response.ok(new GenericEntity<List<Report>>(reports) {})
                       .build();
    }
}

我的实体类:

@Entity
public class Doggy implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @NotNull
    private String dogName;

    public Long getId() {
        return id;
    }

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

    public String getDogName() {
        return dogName;
    }

    public void setDogName(String dogName) {
        this.dogName = dogName;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof Doggy)) {
            return false;
        }
        Doggy other = (Doggy) object;
        if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "com.glasses.pandora.domain.Doggy[ id=" + id + " ]";
    }
}

我的 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="AbcPU" transaction-type="JTA">
    <jta-data-source>java:/jdbc/abc</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>

我在 pom.xml 中的 Maven 依赖项:

<dependencies>
    <dependency>
        <groupId>mysql</groupId>
        <artifactId>mysql-connector-java</artifactId>
        <version>5.1.32</version>
    </dependency>
    <dependency>
        <groupId>org.hibernate</groupId>
        <artifactId>hibernate-entitymanager</artifactId>
        <version>4.3.1.Final</version>
        <scope>provided</scope>
    </dependency>
    <dependency>
        <groupId>unknown.binary</groupId>
        <artifactId>hibernate-jpamodelgen-4.3.1.Final</artifactId>
        <version>SNAPSHOT</version>
        <scope>provided</scope>
    </dependency>
    <dependency>
        <groupId>javax</groupId>
        <artifactId>javaee-web-api</artifactId>
        <version>7.0</version>
        <scope>provided</scope>
    </dependency>
</dependencies>

运行后我的控制台输出:

01:58:54,158 INFO  [stdout] (default task-23) AAAAA Start rs.saveDog();
01:58:54,160 INFO  [stdout] (default task-23) BBBBB Start ReportService.saveDog();
01:58:54,160 INFO  [stdout] (default task-23) BBBBB Persist Success ReportService.saveDog();
01:58:54,160 INFO  [stdout] (default task-23) BBBBB Persist Fail ReportService.saveDog();
01:58:54,161 ERROR [stderr] (default task-23) Error with em.persist(doggy): A JTA EntityManager cannot use getTransaction()
01:58:54,161 INFO  [stdout] (default task-23) BBBBB Stop ReportService.saveDog();
01:58:54,162 INFO  [stdout] (default task-23) AAAAA Stop rs.saveDog();

mysql显示表:

mysql> show tables;
+--------------------+
| Tables_in_abc      |
+--------------------+
| Doggy              |
| hibernate_sequence |
+--------------------+
2 rows in set (0.00 sec)

mysql> SELECT * FROM Doggy;
Empty set (0.00 sec)

mysql>

Answer 1

有几个问题。 hwellmann 提到的第一个是您没有指定数据源。 JTA 需要一个，以便使用与 JTA 事务关联的连接。

第二个是您似乎没有在事务中包装 saveDog 方法。 persist 调用只会在上下文中注册实体；只有在提交关联事务或调用 EntityManager.flush() 时，它才会插入到数据库中。如果您在保存方法中调用 em.flush，这将验证您的 EntityManager 是否正确关联到事务以及插入是否成功。