问题
I'm using hibernate with spring, h2 and liquibase and I'm trying to make a custom String id generator for my entities by taking example with this blog post but I'm getting an error : Caused by: org.hibernate.id.IdentifierGenerationException: Unknown integral data type for ids : java.lang.String
Here my SequenceStyleGenerator code :
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);
}
}
My entity code :
@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;
}
}
And the 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>
Btw is it possible to avoid the parameter sequence_name so hibernate can handle this by itself ?
If anyone can help me, Thanks !
回答1:
The problem is that SequenceStyleGenerator
expects to return a numerical value, not a String
.
I already tried a solution for this problem and it works like a charm. Therefore, you need to change your generator like this:
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;
}
}
Your mapping becomes:
@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;
}
}
Now, when you insert the following entities:
doInJPA(entityManager -> {
entityManager.persist(new Post());
entityManager.persist(new Post("ABC"));
entityManager.persist(new Post());
entityManager.persist(new Post("DEF"));
});
Hibernate generates the right identifier:
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)]
Code available on GitHub.
回答2:
Yes, hibernate have prebuilt String generators. Just substitute your @GenericGenerator
definition to another strategy.
@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;
}
}
For more information about different hibernate generators you can look at documentation.
来源:https://stackoverflow.com/questions/37747218/how-to-implement-a-custom-string-sequence-identifier-generator-with-hibernate