FindByUUID() using Spring Data's JPA Repository

Question

for some reason I have not being able to find a suitable answer for this. I have the following simple entity:

@Entity
@Table(name = \"simple_entity\")
@Acce         


        

Answer 1

I tackled the same question recently, and if someone stumbles up here, the solution for me was to annotate the column with

@Column(name = "id", columnDefinition = "BINARY(16)")
private UUID id;

which solved the problem for me.

Answer 2

Change binary column to String. Default is binary you must add this addnotation

@Type(type="org.hibernate.type.UUIDCharType")

Answer 3

Had the same problem, specifically it works in H2 but not in MySQL.

In addition I got PRIMARY key constraint failures attempting to update the record because Hibernate (under JPA) was querying to see if the record exists and it did not find it.

Using @Column(length=16) also solves this problem neatly, assuming MySQL is using a BINARY column... otherwise the matching will fail due to the column having extra data in the DB (I think it defaults to BINARY[32]).

Answer 4

Try annotate your UUID property with @org.hibernate.annotations.Type(type="org.hibernate.type.UUIDCharType")
or
@org.hibernate.annotations.Type(type="org.hibernate.type.UUIDBinaryType")

I faced a problem similar while query database with UUID due to MSB/LSB swap with UUID in binary format; we solved the problem treating data as String and do necessary conversion before conversion.