英文:
JPA: Override Auto generated ID for composite key?
问题
为了文档目的并检查是否有任何替代解决方案,我有一个实体,其复合键是使用@IdClass定义的。
data class TemplateId(var id: Long? = null, var version: Int = 0) : Serializable@Entity@IdClass(TemplateId::class)data class Template(@Id@Column(name = "id")@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "sequenceGenerator")@SequenceGenerator(name = "sequenceGenerator")var id: Long? = null,@Id@Column(name = "version")var version: Int//其他列)
我的想法是对Template的不同版本使用相同的ID。在插入新模板时,使用序列生成器的效果如预期。但当我尝试插入具有相同ID的新版本行时,@GeneratedValue会覆盖给定的值并自动增加到新值。这里提到的解决方案不起作用。
英文:
Adding this question for documentation purpose, and to check if there is any alternative solution.
I have an entity which has a composite key defined using @IdClass
data class TemplateId(var id: Long? = null, var version: Int = 0) : Serializable@Entity@IdClass(TemplateId::class)data class Template(@Id@Column(name = "id")@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "sequenceGenerator")@SequenceGenerator(name = "sequenceGenerator")var id: Long? = null,@Id@Column(name = "version")var version: Int//Other columns)
Idea is to have same ID for different versions of the Template. Using Sequence generator works as expected while inserting new template.
But when I try to insert a new version row with same ID, @GeneratedValue overrides the given value and autoincrements to new value.
Solution mentioned in JPA: Override Auto generated ID
does not work.
答案1
得分: 0
我尝试了以下选项,但都没有成功。
-
如问题中所述,无法使用
@GeneratedValue,因为它会替换给定的值。 -
不能用自定义生成器(
@GenericGenerator)替换@SequenceGenerator,因为它在复合键中不起作用。它试图将 Long 值转换为 IdClassTemplateId。 -
我正在使用 Postgres,所以尝试将
id的列类型设置为SERIAL。但是这在复合键上使用 IDENTITY GenerationType 时不起作用。存在一个现有的问题:HHH-9662。 -
不能使用带有 NULL 值的数据库自增,Postgres 会报违反约束的错误。
-
"insertable"=false / "updatable"=false在id列上不适用于@Id列。 -
类似地,尝试使用 Hibernate 的
@DynamicInsert,这样它将在插入查询中跳过空列值,但即使在@Id列中也不起作用。
最终不得不覆盖 Spring 的 JpaRepository 的 save 函数,使其工作起来。
interface CustomTemplateRepository<S> {fun <E: S> save(entity: E): Efun <E: S> saveAndFlush(entity: E): E}class TemplateRepositoryImpl(private val jdbcTemplate: NamedParameterJdbcTemplate,@PersistenceContext private val entityManager: EntityManager) : CustomTemplateRepository<Template> {override fun <E : Template> save(entity: E): E {if (entity.id == null)entity.id = jdbcTemplate.queryForObject("select nextval('sequence_generator')",mutableMapOf<String, Any>(), Long::class.java)entityManager.persist(entity)return entity}override fun <E : Template> saveAndFlush(entity: E): E {return save(entity).also {entityManager.flush()}}}
英文:
I tried out following options and none worked.
-
As mentioned in the question, cannot use
@GeneratedValueas it replaces the given value -
Cannot replace
@SequenceGeneratorwith custom generator (@GenericGenerator), doesn't work with composite key. It tries to cast Long value to IdClassTemplateId -
I am using Postgres, so tried using column type
SERIALforid. This does not work with IDENTITY GenerationType on composite key. There is an existing issue : HHH-9662 -
Cannot use DB auto-increment with NULL value, postgres gives constraint violation error
-
"insertable"=false/ "updatable"=falseonidcolumn does not work for@Idcolumns -
Similarly tried using hibernate's
@DynamicInsertso that it will skip null column values in insert query, but even that doesn't work with@Id
Finally had to override the save function of Spring's JpaRepository to make it work
interface CustomTemplateRepository<S> {fun <E: S> save(entity: E): Efun <E: S> saveAndFlush(entity: E): E}class TemplateRepositoryImpl(private val jdbcTemplate: NamedParameterJdbcTemplate,@PersistenceContext private val entityManager: EntityManager) : CustomTemplateRepository<Template> {override fun <E : Template> save(entity: E): E {if(entity.id == null)entity.id = jdbcTemplate.queryForObject("select nextval('sequence_generator')",mutableMapOf<String, Any>(), Long::class.java)entityManager.persist(entity)return entity}override fun <E : Template> saveAndFlush(entity: E): E {return save(entity).also {entityManager.flush()}}}
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