6

我正在使用带有 spring、h2 和 liquibase 的 hibernate，并且我正在尝试通过以这篇博客文章为例，为我的实体制作一个自定义的 String id 生成器，但我遇到了一个错误：Caused by: org.hibernate.id.IdentifierGenerationException: Unknown integral data type for ids : java.lang.String

这是我的 SequenceStyleGenerator 代码：

public class CTCIDGenerator extends SequenceStyleGenerator {

    @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;
            }
        }
        return "CTC"+super.generate(session, obj);
    }
}

我的实体代码：

@Entity
@Table(name = "contact")
public class Contact implements Serializable, Identifiable<String> {

    private static final long serialVersionUID = 1L;

    @Id
    @GenericGenerator(
        name = "assigned-sequence",
        strategy =     "net.atos.seirich.support.domain.idgenerator.CTCIDGenerator",
        parameters = @org.hibernate.annotations.Parameter(
            name = "sequence_name", 
            value = "hibernate_sequence"
        )
    )
    @GeneratedValue(generator = "assigned-sequence", strategy = GenerationType.SEQUENCE)
    private String id;

    public String getId() {
        return id;
    }

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

还有 liquibase XML：

<?xml version="1.0" encoding="utf-8"?>
<databaseChangeLog
xmlns="http://www.liquibase.org/xml/ns/dbchangelog"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.liquibase.org/xml/ns/dbchangelog http://www.liquibase.org/xml/ns/dbchangelog/dbchangelog-3.4.xsd">

    <property name="autoIncrement" value="true" dbms="mysql,h2,postgresql,oracle"/>

    <property name="floatType" value="float4" dbms="postgresql, h2"/>
    <property name="floatType" value="float" dbms="mysql, oracle"/>

    <changeSet id="20160513091901-1" author="jhipster">
        <createTable tableName="contact">
            <column name="id" type="longvarchar" autoIncrement="${autoIncrement}">
                <constraints primaryKey="true" nullable="false"/>
            </column>
    </changeSet>
</databaseChangeLog>

顺便说一句，是否可以避免参数 sequence_name 以便休眠可以自己处理？

如果有人可以帮助我，谢谢！

4

2 回答 2

12

问题是SequenceStyleGenerator期望返回一个数值，而不是一个String.

我已经尝试过解决这个问题，它就像一个魅力。因此，您需要像这样更改生成器：

public class StringSequenceIdentifier implements IdentifierGenerator, Configurable {

    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();

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

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

        sequenceCallSyntax = dialect.getSequenceNextValString(ConfigurationHelper.getString(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 "CTC" + seqValue;
    }
}

您的映射变为：

@Entity(name = "Post")
@Table(name = "post")
public static 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")
    )
    @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;
    }
}

现在，当您插入以下实体时：

doInJPA(entityManager -> {
    entityManager.persist(new Post());
    entityManager.persist(new Post("ABC"));
    entityManager.persist(new Post());
    entityManager.persist(new Post("DEF"));
});

Hibernate 生成正确的标识符：

Query:["select nextval ('hibernate_sequence')"], Params:[()]
Query:["select nextval ('hibernate_sequence')"], Params:[()]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, CTC1)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, ABC)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, CTC2)]
Query:["insert into post (version, id) values (?, ?)"], Params:[(0, DEF)]

代码在GitHub上可用。

于 2016-06-10T13:29:27.520 回答
0

是的，hibernate 有预建的字符串生成器。只需将您的定义替换@GenericGenerator为另一种策略即可。

@Entity
@Table(name = "contact")
public class Contact implements Serializable, Identifiable<String> {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(generator = "uuid")
    @GenericGenerator(name = "uuid", strategy = "uuid2")
    private String id;

    public String getId() {
        return id;
    }

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

有关不同休眠生成器的更多信息，您可以查看文档。

于 2016-06-10T11:58:06.757 回答