java - 使用多个@GenericGenerator？

Question

我在 Spring Boot 项目 (Hibernate 5) 中尝试了 Hibernate 的 @GenericGenerator。制作了这样的测试实体:

@Entity
public class BatchTest implements Serializable {
    private static final long serialVersionUID = 3012542467060581674L;

    @Id
    @GeneratedValue(generator = "batchTestIdGenerator")
    @GenericGenerator(
            name = "batchTestIdGenerator",
            strategy = "org.hibernate.id.enhanced.SequenceStyleGenerator",
            parameters = {
                    @org.hibernate.annotations.Parameter(name = "increment_size",value = "1000")
            }
    )
    private long id;

    @Lob
    @Column(nullable = false)
    private String someVal;
...

使用这个生成器，我可以使用 JPA 快速插入实体(我有一个 Spring 存储库，由 @Service 提供服务)，这很酷，但是序列实际上是如何存储的？我的数据库当前是一个 SQL Server 2016 实例，在架构中，我可以看到我有一个 dbo.hibernate_sequence :

如果我使用不同的生成器创建第二个实体，我会在启动过程中遇到异常:

@Id
@GeneratedValue(generator = "batchTestIdGenerator2", strategy = GenerationType.SEQUENCE)
@GenericGenerator(
        name = "batchTestIdGenerator2",
        strategy = "org.hibernate.id.enhanced.SequenceStyleGenerator",
        parameters = {
                @org.hibernate.annotations.Parameter(name = "increment_size",value = "500")
        }
)
private long id;

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaAutoConfiguration.class]: Invocation of init method failed; nested exception is org.hibernate.HibernateException: Multiple references to database sequence [hibernate_sequence] were encountered attempting toset conflicting values for 'increment size'.  Found [500] and [1000]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1628) ~[spring-beans-4.3.12.RELEASE.jar:4.3.12.RELEASE]

我可以定义不同的序列吗？看起来它尝试对两个序列使用相同的 dbo.hibernate_sequence (id 没有在数据库中创建新的 dbo 序列)并且只是忽略名称。如果可以定义N个序列，我该怎么做？

编辑:我尝试手动创建序列(匹配名称)，但仍然遇到相同的错误。

Answer 1

Hibernate 允许您创建自己的序列生成器，它允许您使用 Hibernate IdentifierGenerator 定义格式和表格。请尝试以下方法。

通过实现 IdentifierGenerator 创建一个 SequenceGenerator 类，使用 org.hibernate.id.Configurable 接口(interface)使您的 Generator 可配置 -它将接受来自实体类的参数-

自定义标识符生成器如下所示:

public class StringSequenceIdentifier 
        implements IdentifierGenerator, Configurable {

    public static final String SEQUENCE_PREFIX = "sequence_prefix";

    private String sequencePrefix;

    private String sequenceCallSyntax;

    @Override
    public void configure(
            Type type, Properties params, ServiceRegistry serviceRegistry)
            throws MappingException {
        final JdbcEnvironment jdbcEnvironment =
                serviceRegistry.getService(JdbcEnvironment.class);
        final Dialect dialect = jdbcEnvironment.getDialect();

        sequencePrefix = ConfigurationHelper.getString(
                SEQUENCE_PREFIX,
                params,
                "SEQ_");

        final String sequencePerEntitySuffix = ConfigurationHelper.getString(
                SequenceStyleGenerator.CONFIG_SEQUENCE_PER_ENTITY_SUFFIX,
                params,
                SequenceStyleGenerator.DEF_SEQUENCE_SUFFIX);

        final String defaultSequenceName = ConfigurationHelper.getBoolean(
                SequenceStyleGenerator.CONFIG_PREFER_SEQUENCE_PER_ENTITY,
                params,
                false)
                ? params.getProperty(JPA_ENTITY_NAME) + sequencePerEntitySuffix
                : SequenceStyleGenerator.DEF_SEQUENCE_NAME;

        sequenceCallSyntax = dialect.getSequenceNextValString(
                ConfigurationHelper.getString(
                        SequenceStyleGenerator.SEQUENCE_PARAM,
                        params,
                        defaultSequenceName));
    }

    @Override
    public Serializable generate(SessionImplementor session, Object obj) {
        if (obj instanceof Identifiable) {
            Identifiable identifiable = (Identifiable) obj;
            Serializable id = identifiable.getId();
            if (id != null) {
                return id;
            }
        }
        long seqValue = ((Number) Session.class.cast(session)
            .createSQLQuery(sequenceCallSyntax)
            .uniqueResult()).longValue();

        return sequencePrefix + String.format("%011d%s", 0 ,seqValue);
    }
}

实体会像

@Entity(name = "Post") @Table(name = "post")
public class Post implements Identifiable<String> {

    @Id
    @GenericGenerator(
        name = "assigned-sequence",
        strategy = "com.vladmihalcea.book.hpjp.hibernate.identifier.StringSequenceIdentifier",
        parameters = {
            @org.hibernate.annotations.Parameter(
                name = "sequence_name", value = "hibernate_sequence"),
            @org.hibernate.annotations.Parameter(
                name = "sequence_prefix", value = "CTC_"),
        }
    )
    @GeneratedValue(
        generator = "assigned-sequence", 
        strategy = GenerationType.SEQUENCE)
    private String id;

    @Version
    private Integer version;

    public Post() {
    }

    public Post(String id) {
        this.id = id;
    }

    @Override
    public String getId() {
        return id;
    }
}

详细说明请引用以下链接。 https://vladmihalcea.com/how-to-implement-a-custom-string-based-sequence-identifier-generator-with-hibernate/