Hibernate委托将UUID生成任务交给PostgreSQL。

huangapple
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英文:

Hibernate to delegate UUID generation to postgres

问题

I understand that you want the translation of the code portions from your provided text. Here they are:

  1. <databaseChangeLog
  2.         xmlns="http://www.liquibase.org/xml/ns/dbchangelog"
  3.         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  4.         xsi:schemaLocation="http://www.liquibase.org/xml/ns/dbchangelog http://www.liquibase.org/xml/ns/dbchangelog/dbchangelog-latest.xsd">
  5.  
  6.     <changeSet author="..." id="uuid-ossp-extension">
  7.         <sql>CREATE EXTENSION IF NOT EXISTS "uuid-ossp"</sql>
  8.     </changeSet>
  9.  
  10.     <changeSet author="example" id="widgets-table">
  11.         <createTable tableName="widgets">
  12.             <column name="widget_id" type="uuid" defaultValueComputed="uuid_generate_v1mc()">
  13.                 <constraints nullable="false" primaryKey="true" primaryKeyName="widget_pk"/>
  14.             </column>
  15.             <column name="name" type="varchar(10)">
  16.                 <constraints nullable="false"/>
  17.             </column>
  18.         </createTable>
  19.     </changeSet>
  20.  
  21. </databaseChangeLog>
  1. @Repository
  2. @RequiredArgsConstructor
  3. public class WidgetRepository {
  5.     public static final String INSERT_QUERY = "INSERT INTO widgets(name) VALUES (:name)";
  8.     private final NamedParameterJdbcTemplate template;
  10.     public UUID saveAndReturn(Widget widget) {
  11.         var params = new MapSqlParameterSource(Map.of("name", widget.getName());
  12.         var generatedKeyHolder = new GeneratedKeyHolder();
  13.         var generateKeyColumns = new String[]{"widget_id"};
  14.         template.update(INSERT_QUERY, params, generatedKeyHolder, generateKeyColumns);
  15.         return generatedKeyHolder.getKeyAs(UUID.class);
  16.     }
  18.     ...
  20. }
  1. @Entity
  2. @Table(name = "widgets")
  3. public class Widget {
  4.  
  5.     @Id
  6.     @UuidGenerator(style = TIME)
  7.     @Column(name = "widget_id")
  8.     private UUID id;
  9.  
  10.     @Column
  11.     private String name;
  13.     ...
  14. }

Please note that I've replaced some special characters used for HTML/XML encoding in your provided code with their respective characters for readability.

英文:

Imagine I have a simple table such as the following, defined using a liquibase changeset

  1. <databaseChangeLog
  2.         xmlns="http://www.liquibase.org/xml/ns/dbchangelog"
  3.         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  4.         xsi:schemaLocation="http://www.liquibase.org/xml/ns/dbchangelog http://www.liquibase.org/xml/ns/dbchangelog/dbchangelog-latest.xsd">
  5.  
  6.     <changeSet author="..." id="uuid-ossp-extension">
  7.         <sql>CREATE EXTENSION IF NOT EXISTS "uuid-ossp"</sql>
  8.     </changeSet>
  9.  
  10.     <changeSet author="example" id="widgets-table">
  11.         <createTable tableName="widgets">
  12.             <column name="widget_id" type="uuid" defaultValueComputed="uuid_generate_v1mc()">
  13.                 <constraints nullable="false" primaryKey="true" primaryKeyName="widget_pk"/>
  14.             </column>
  15.             <column name="name" type="varchar(10)">
  16.                 <constraints nullable="false"/>
  17.             </column>
  18.         </createTable>
  19.     </changeSet>
  20.  
  21. </databaseChangeLog>

That is, I want to be explicit about letting postgres compute the UUID.

If I was using Spring JdbcTemplate, I could use something simple like:

  1. @Repository
  2. @RequiredArgsConstructor
  3. public class WidgetRepository {
  5.     public static final String INSERT_QUERY = "INSERT INTO widgets(name) VALUES (:name)";
  8.     private final NamedParameterJdbcTemplate template;
  10.     public UUID saveAndReturn(Widget widget) {
  11.         var params = new MapSqlParameterSource(Map.of("name", widget.getName());
  12.         var generatedKeyHolder = new GeneratedKeyHolder();
  13.         var generateKeyColumns = new String[]{"widget_id"};
  14.         template.update(INSERT_QUERY, params, generatedKeyHolder, generateKeyColumns);
  15.         return generatedKeyHolder.getKeyAs(UUID.class);
  16.     }
  18.     ...
  20. }

With Hibernate 6.0 onwards, there is a new UuidGenerator annotation, that simplifies Hibernate taking responsibility for generating a UUID:

  1. @Entity
  2. @Table(name = "widgets")
  3. public class Widget {
  4.  
  5.     @Id
  6.     @UuidGenerator(style = TIME)
  7.     @Column(name = "widget_id")
  8.     private UUID id;
  9.  
  10.     @Column
  11.     private String name;
  13.     ...
  14. }

However, this new UUIDGenerator mechanism doesn't seem to support delegating to postgres. https://vladmihalcea.com/uuid-identifier-jpa-hibernate/ suggested an alternative approach, but the UUIDGenerationStrategy it implements is now deprecated.

Does anyone know the preferred mechanism for Hibernate 6.0 onwards?

答案1

得分: 2

I tweeted about this here:

<https://twitter.com/1ovthafew/status/1645795161379831811>

The short answer is:

  1. @Entity
  2. public class Entity {
  3.     @Generated @Id
  4.     @ColumnDefault(&quot;gen_random_uuid()&quot;)
  5.     UUID uuid;
  6. }

Note that Hibernate 6.2 is required.

英文:

I tweeted about this here:

<https://twitter.com/1ovthafew/status/1645795161379831811>

The short answer is:

  1. @Entity
  2. public class Entity {
  3.     @Generated @Id
  4.     @ColumnDefault(&quot;gen_random_uuid()&quot;)
  5.     UUID uuid;
  6. }

Note that Hibernate 6.2 is required.

huangapple
  • 本文由 发表于 2023年5月10日 16:09:01
  • 转载请务必保留本文链接：https://go.coder-hub.com/76216222.html
  • hibernate
  • jpa
  • postgresql
  • spring
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