龙目岛对jpa有副作用吗

Question

我正在将一个jpa实体转换为使用Lombok。结果代码如下：

@Entity
@Table(name = "TEST")
@Data
@NoArgsConstructor
@AllArgsConstructor
class Test {
   ...
   @Column(name = "FORMATTING")
   @Enumerated(EnumType.ORDINAL)
   private FormatType formatType;
   ...
}

结果错误消息包含以下内容

Caused by: org.hibernate.HibernateException: Missing column: formatType in TEST

13:22:19,967 INFO  [org.hibernate.tool.hbm2ddl.TableMetadata] (ServerService Thread Pool -- 57) HHH000261: Table found: TABLE
13:22:19,967 INFO  [org.hibernate.tool.hbm2ddl.TableMetadata] (ServerService Thread Pool -- 57) HHH000037: Columns: [..., formatType, ...]
13:22:19,968 ERROR [org.jboss.msc.service.fail] (ServerService Thread Pool -- 57) MSC000001: Failed to start service jboss.persistenceunit."...": org.jboss.msc.service.StartException in service jboss.persistenceunit."...": javax.persistence.PersistenceException: [PersistenceUnit: ...] Unable to build EntityManagerFactory

  @Entity
  @Inheritance(strategy = InheritanceType.JOINED)
  @Table(name = "PARENT")

  public abstract class Parent implements Serializable {

     private static final long serialVersionUID = 1;
     @Id
     @Column(name = "ID")
     @GeneratedValue
     private long id;
     @Column(name = "ENABLED")
     private boolean enabled;
  }

  @Entity
  @Table(name = "CHILD")
  @Data
  @NoArgsConstructor
  @AllArgsConstructor
  public class Child extends Parent {
     /** XXX: HERE BE DRAGONS */
     @Column(name = "ENUM_1")
     @Enumerated(EnumType.STRING)
     private Enum1 enum1;
     @Column(name = "ENUM_2")
     @Enumerated(EnumType.ORDINAL)
     private Enum2 enum2;
     /** XXX: NO MORE DRAGONS */
     @Column(name = "FREQUENCY")
     private String frequency;
     @Column(name = "EMPTY")
     private boolean empty;
     @Column(name = "MAX_SIZE")
     private int maxSize;
  }
  public enum Enum1 {
     A,
     B,
     C
  }
  public enum Enum2 {
     X,
     Y,
     Z
  }

我已经回滚了龙目岛的变化，我仍然想知道是什么问题，但没有匆忙。另外，由于这个可爱的小虫子，我大约落后了4个小时，所以我可能会有点慢的反应。

子表的pk是父表的fk，如果没有lombok，一切似乎都可以工作，尽管child类没有ID。

解决方案：

@Entity
@Table(name = "TEST")
@Setter
@Getter
class Test {
   ...
   @Column(name = "FORMATTING")
   @Enumerated(EnumType.ORDINAL)
   private FormatType formatType;
   ...
}

@Entity
@Table(name = "TEST")
class Test {
   ...
   @Column(name = "FORMATTING")
   @Enumerated(EnumType.ORDINAL)
   private FormatType formatType;

   public FormatType getFormatType() {
      return formatType;
   }
   public void setFormatType(FormatType formatType) {
      this.formatType = formatType;
   }
   ...
}

@Entity
@Table(name = "TEST")
class Test {
   ...
   private FormatType formatType;

   @Column(name = "FORMATTING")
   @Enumerated(EnumType.ORDINAL)
   public FormatType getFormatType() {
      return formatType;
   }
   public void setFormatType(FormatType formatType) {
      this.formatType = formatType;
   }
   ...
}

在经历了大量的困惑之后，我试图找到一些示例，并填写一些关于龙目岛是如何执行其操作的细节（平心而论，我很容易感到困惑），我发现了这个小宝石:onmethod=@__（{@annotationshere})。利用这一特性，我提出了以下内容：

@Entity
@Table(name = "TEST")
@Setter
class Test {
   ...
   @Getter(onMethod=@__({
         @Column(name = "FORMATTING"),
         @Enumerated(EnumType.ORDINAL)
      }))
   private FormatType formatType;

   ...
}

很快就能奏效了。现在我们已经有了显然是唯一可用的解决方案，我想解决我们现在都在思考的问题：这真的比手动编写方法并在那里附加注释更干净吗？回答:……我不知道。我很高兴我找到了解决办法。

Answer 1

真奇怪。你能显示更多的代码吗？我试图用你的问题中的部分代码来编写一个简单的项目，结果成功了。我使用了Spring Boot和MySQL。请尝试检查您的配置。这是我的代码：

枚举：

public enum FormatType {

    FIRST_TYPE, SECOND_TYPE
}

MySQL中的表：

create table TEST
(
    ID int auto_increment primary key,
    FORMATTING int not null
);

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;

import javax.persistence.*;

@Entity
@Table(name = "TEST")
@Data
@NoArgsConstructor
@AllArgsConstructor
public class Test {

    @Id
    @Column(name = "ID")
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer id;

    @Column(name = "FORMATTING")
    @Enumerated(EnumType.ORDINAL)
    private FormatType formatType;
}

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;

@Repository
public interface TestRepository extends JpaRepository<Test, Integer> {
}

服务：

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

import java.util.List;

@Service
public class TestService {

    private TestRepository repository;

    @Autowired
    public TestService(TestRepository repository) {
        this.repository = repository;
    }

    public List<Test> getAllTestEntities() {
        return repository.findAll();
    }
}