在jpa中调用序列的下一个值

I have a class mapped as an Entity to persist it in a database. I have an id field as the primary key so every time the object is persisted the value of the id is retrieved from the sequence "myClass_pk_seq", a code like the following one.

我有一个类映射为实体,以将其持久保存在数据库中。我有一个id字段作为主键,所以每次持久化对象时,id的值都从序列“myClass_pk_seq”中检索,代码如下所示。

@Entity
@Table(name="myObjects")
public class MyClass {
    @Id
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="sequence")
    @SequenceGenerator(name="sequence", sequenceName="myClass_pk_seq", allocationSize=1)
    @Column(name="myClassId")
    private Integer id;

    private Integer code;


    ...
}

I need to make in "code" attribute something similar to id. I need to have a sequence so I can assign to code the next value of the sequence (to reserve that value in case the user wants to reserve it but without persisting data). I mean the user will see the field, if he doesn't know what to enter he can push a button and receieve in the screen the next possible value (and he can or not accept it). How can I get the next value of a sequence defined in JPA (and increment its value) without persisting the data for a non primary key field?

我需要在“code”属性中创建类似于id的东西。我需要有一个序列,所以我可以为代码分配序列的下一个值(如果用户想要保留它但不保留数据,则保留该值)。我的意思是用户将看到该字段,如果他不知道输入什么,他可以按下按钮并在屏幕上接收下一个可能的值(并且他可以接受或不接受它)。如何在不保留非主键字段的数据的情况下获取JPA中定义的序列的下一个值(并递增其值)?

I just want to have a method which call nextval on a sequence associated with "code" field, and returns the value. What's the best way to do it in JPA with annotations?

我只想要一个方法在与“code”字段关联的序列上调用nextval,并返回该值。使用注释在JPA中执行此操作的最佳方法是什么?

Thanks.

2 个解决方案

#1

I just want to have a method which call nextval on a sequence associated with "code" field, and returns the value. What's the best way to do it in JPA with annotations?

我只想要一个方法在与“code”字段关联的序列上调用nextval,并返回该值。使用注释在JPA中执行此操作的最佳方法是什么?

Use native SQL to get the next sequence value when the user pushes the button. Either create the sequence manually or use a "fake entity" to have JPA create it for you.

当用户按下按钮时,使用本机SQL获取下一个序列值。手动创建序列或使用“假实体”让JPA为您创建序列。

If you don't want to use native SQL, insert an entity relying on the sequence and gets its id.

如果您不想使用本机SQL,请插入依赖于序列的实体并获取其ID。

Both solutions sounds a bit ugly. Maybe you could simply use a random generator like a UUID generator.

两种解决方案听起来都有些难看。也许你可以简单地使用像UUID生成器这样的随机生成器。

Actually, you didn't mention anything about the uniqueness of the code (and the JPA annotations don't show it must be unique). Why don't you return a random int?

实际上,您没有提及有关代码唯一性的任何内容(并且JPA注释未显示它必须是唯一的)。你为什么不返回一个随机的int?

#2

See another question/answers on the subject of using sequence defined elsewhere than id fields. You can create a fake entity with one field (of type Long id). Connect it to the sequence you defined in the DB. Then create a CrudRepository implementation for that entity and call its save() method with an empty instance of the fake entity object you defined. Hibernate will run for you a "select YOUR_SEQ.NEXTVAL from dual" query.

请参阅有关使用除id字段以外定义的序列的主题的其他问题/答案。您可以使用一个字段(Long id类型)创建虚假实体。将其连接到您在DB中定义的序列。然后为该实体创建一个CrudRepository实现,并使用您定义的假实体对象的空实例调用其save()方法。 Hibernate将为您运行“从双重选择YOUR_SEQ.NEXTVAL”查询。

#1