How to set autogenerated Id manually?

Question

I have an entity called Task and it has taskId field. Problem is I need to create/update some specific tasks and JPA autogenerator doesn't allow me to set taskId manually.(It overrides my taskId)

Is there any way to set taskId manually?

@Id
@Column(name = "task_id")
@GeneratedValue(generator = "system-uuid")
@GenericGenerator(name = "system-uuid", strategy = "org.hibernate.id.UUIDGenerator")
private String taskId;

The JPA spec doesn't allow for a user wanting to set under some situation and generate under others (since the definition of `GeneratedValue` is static). Some JPA providers do support it (DataNucleus does IIRC) but non-portable if relying on it — Neil Stockton, Jul 14 '17 at 12:45
Does this answer your question? [set Id (PK) generation value auto and manual](https://stackoverflow.com/questions/18182826/set-id-pk-generation-value-auto-and-manual) — miken32, Aug 13 '22 at 16:19

score 3 · Accepted Answer · edited Jul 30 '18 at 08:19

3

Using

@GeneratedValue(strategy = GenerationType.IDENTITY)

Example: http://www.codejava.net/frameworks/hibernate/java-hibernate-jpa-annotations-tutorial-for-beginners

edited Jul 30 '18 at 08:19

Bugs

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answered Jul 30 '18 at 04:09

Quockhanh Pham

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score 2 · Answer 2 · answered Jul 15 '17 at 09:34

@GeneratedValue annotation doesn't allow the user to set the value manually.Instead of using that annotation you can manually generate some string format like below and mark the taskId column with @Id

String hql = "from MasterCount where itemType=:type";
Query<MasterCount> query = session.createQuery(hql);
query.setParameter("type", "task");
MasterCount count = query.uniqueResult();
int lastId = count.getItemId() + 1;
count.setItemId(lastId);
task.setTaskId("task0"+lastId);

That is create a master table with columns itemId and itemType.Save "task" and 0 in the columns respectively.Check the last entered value ,increment one and save again.

score 2 · Answer 3 · answered Jul 16 '17 at 14:17

2

you can initialize it manually.

        @Id
        @Column(name= "id")
        @GeneratedValue
        private Long id = Long.parseLong(UUID.randomUUID().toString(), 16);

answered Jul 16 '17 at 14:17

dhyanandra singh

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But here it will go to update entity insted of insert because id is not null in jpa.save() method – Dharmendrasinh Chudasama Aug 13 '22 at 12:37

score 1 · Answer 4 · answered Jul 14 '17 at 12:22

1

This link may help you.

I would apply a basic solution if I understood your problem correctly.

@Column(name = "task_id", unique = true, nullable = false)
private String taskId;

answered Jul 14 '17 at 12:22

memoricab

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The link mentioned in the answer helps! – Hong Jun 30 '22 at 07:10

score 0 · Answer 5 · answered Aug 13 '22 at 16:08

Just implement org.springframework.data.domain.Persistable;

After too much finding and frustration i have used this and worked for my R2DBMS entity

@Table(name = "...")
public class .... implements Persistable<String> {

   ...

   public boolean isNew(){
     return id==null; //modify and set your logic
   }

}

How to set autogenerated Id manually?

5 Answers5