PostgreSQL枚举和Java枚举之间的Hibernate映射

Question

背景

• Spring 3.x,JPA 2.0,Hibernate 4.x,Postgresql 9.x.
• 使用我要映射到Postgresql枚举的枚举属性处理Hibernate映射类.

问题

查询枚举列上的where子句会引发异常.

org.hibernate.exception.SQLGrammarException: could not extract ResultSet
... 
Caused by: org.postgresql.util.PSQLException: ERROR: operator does not exist: movedirection = bytea
  Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts.

代码(大大简化)

SQL:

create type movedirection as enum (
    'FORWARD', 'LEFT'
);

CREATE TABLE move
(
    id serial NOT NULL PRIMARY KEY,
    directiontomove movedirection NOT NULL
);

Hibernate映射类:

@Entity
@Table(name = "move")
public class Move {

    public enum Direction {
        FORWARD, LEFT;
    }

    @Id
    @Column(name = "id")
    @GeneratedValue(generator = "sequenceGenerator", strategy=GenerationType.SEQUENCE)
    @SequenceGenerator(name = "sequenceGenerator", sequenceName = "move_id_seq")
    private long id;

    @Column(name = "directiontomove", nullable = false)
    @Enumerated(EnumType.STRING)
    private Direction directionToMove;
    ...
    // getters and setters
}

调用查询的Java:

public List<Move> getMoves(Direction directionToMove) {
    return (List<Direction>) sessionFactory.getCurrentSession()
            .getNamedQuery("getAllMoves")
            .setParameter("directionToMove", directionToMove)
            .list();
}

Hibernate xml查询:

<query name="getAllMoves">
    <![CDATA[
        select move from Move move
        where directiontomove = :directionToMove
    ]]>
</query>

故障排除

• 查询id而不是枚举按预期工作.

• 没有数据库交互的Java工作正常:

  public List<Move> getMoves(Direction directionToMove) {
      List<Move> moves = new ArrayList<>();
      Move move1 = new Move();
      move1.setDirection(directionToMove);
      moves.add(move1);
      return moves;
  }
  

• createQuery而不是使用XML进行查询,类似于Apache的JPA和Enums中的findByRating示例,通过@Enumerated文档给出了相同的异常.
• 使用select * from move where direction = 'LEFT';预期的工作查询psql .
• XML where direction = 'FORWARD'中的查询中的硬编码有效.

• .setParameter("direction", direction.name())不,同样与.setString()和.setText(),异常的变化:

  Caused by: org.postgresql.util.PSQLException: ERROR: operator does not exist: movedirection = character varying
  

试图解决

• UserType按照此接受的答案建议自定义/sf/answers/111581431/以及:

  @Column(name = "direction", nullable = false)
  @Enumerated(EnumType.STRING) // tried with and without this line
  @Type(type = "full.path.to.HibernateMoveDirectionUserType")
  private Direction directionToMove;
  

• 如上所述,EnumType通过更高评级但未被接受的答案/sf/answers/112300051/建议使用Hibernate映射,以及:

  @Type(type = "org.hibernate.type.EnumType",
      parameters = {
              @Parameter(name  = "enumClass", value = "full.path.to.Move$Direction"),
              @Parameter(name = "type", value = "12"),
              @Parameter(name = "useNamed", value = "true")
      })
  

  在看到/sf/answers/926898731/之后,有和没有两个第二个参数

• 尝试在这个答案/sf/answers/1417655081/中注释getter和setter .
• 没有尝试,EnumType.ORDINAL因为我想坚持EnumType.STRING,这不那么脆弱,更灵活.

其他说明

JPA 2.1类型转换器不是必需的,但不管怎样都不是,因为我现在正在使用JPA 2.0.

Answer 1

您不必手动创建以下所有Hibernate类型.您可以使用以下依赖项通过Maven Central获取它们:

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-52</artifactId>
    <version>${hibernate-types.version}</version>
</dependency>

有关更多信息,请查看hibernate类型的开源项目.

正如我在本文中所解释的,如果您使用以下自定义类型轻松地将Java Enum映射到PostgreSQL Enum列类型:

public class PostgreSQLEnumType extends org.hibernate.type.EnumType {

    public void nullSafeSet(
            PreparedStatement st, 
            Object value, 
            int index, 
            SharedSessionContractImplementor session) 
        throws HibernateException, SQLException {
        if(value == null) {
            st.setNull( index, Types.OTHER );
        }
        else {
            st.setObject( 
                index, 
                value.toString(), 
                Types.OTHER 
            );
        }
    }
}

要使用它,您需要使用Hibernate @Type注释来注释该字段,如以下示例所示:

@Entity(name = "Post")
@Table(name = "post")
@TypeDef(
    name = "pgsql_enum",
    typeClass = PostgreSQLEnumType.class
)
public static class Post {

    @Id
    private Long id;

    private String title;

    @Enumerated(EnumType.STRING)
    @Column(columnDefinition = "post_status_info")
    @Type( type = "pgsql_enum" )
    private PostStatus status;

    //Getters and setters omitted for brevity
}

此映射假定您post_status_info在PostgreSQL中具有枚举类型:

CREATE TYPE post_status_info AS ENUM (
    'PENDING', 
    'APPROVED', 
    'SPAM'
)

就是这样,它就像一个魅力.这是对GitHub的测试证明了这一点.

Answer 2

HQL

正确别名并使用限定属性名称是解决方案的第一部分.

<query name="getAllMoves">
    <![CDATA[
        from Move as move
        where move.directionToMove = :direction
    ]]>
</query>

Hibernate映射

@Enumerated(EnumType.STRING)仍然没有工作,所以一个习惯UserType是必要的.关键是要正确覆盖,nullSafeSet如在这个答案/sf/answers/533024971/和网络上的类似 实现.

@Override
public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor session) throws HibernateException, SQLException {
    if (value == null) {
        st.setNull(index, Types.VARCHAR);
    }
    else {
        st.setObject(index, ((Enum) value).name(), Types.OTHER);
    }
}

车辆改道

implements ParameterizedType 没合作:

org.hibernate.MappingException: type is not parameterized: full.path.to.PGEnumUserType

所以我无法像这样注释枚举属性:

@Type(type = "full.path.to.PGEnumUserType",
        parameters = {
                @Parameter(name = "enumClass", value = "full.path.to.Move$Direction")
        }
)

相反,我宣布这样的类是这样的:

public class PGEnumUserType<E extends Enum<E>> implements UserType

使用构造函数:

public PGEnumUserType(Class<E> enumClass) {
    this.enumClass = enumClass;
}

遗憾的是,这意味着任何其他类似映射的枚举属性都需要这样的类:

public class HibernateDirectionUserType extends PGEnumUserType<Direction> {
    public HibernateDirectionUserType() {
        super(Direction.class);
    }
}

注解

注释财产,你就完成了.

@Column(name = "directiontomove", nullable = false)
@Type(type = "full.path.to.HibernateDirectionUserType")
private Direction directionToMove;

其他说明

• EnhancedUserType 以及它想要实施的三种方法

  public String objectToSQLString(Object value)
  public String toXMLString(Object value)
  public String objectToSQLString(Object value)
  

  我没看到任何差别,所以我坚持了下来implements UserType.

• 根据您使用该类的方式,可能没有必要通过覆盖nullSafeGet两个链接解决方案的方式来使其特定于postgres .
• 如果您愿意放弃postgres枚举,您可以使列text和原始代码无需额外工作.