I am using Oracle database and have to define a UUID column. I followed other posts and created the table using regex pattern for primary key:

CREATE TABLE TEST_UUID (
    ID_UUID VARCHAR(255)
    DEFAULT REGEXP_REPLACE(RAWTOHEX(SYS_GUID()), '([A-F0-9]{8})([A-F0-9]{4})([A-F0-9]{4})([A-F0-9]{4})([A-F0-9]{12})', '\1-\2-\3-\4-\5'),
    NAME VARCHAR2(40)  NOT NULL
);

In my entity class I have to define the ID_UUID column of type UUID (cannot define it as String). The entity class looks like this:

import java.util.UUID;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "test_uuid")
public class TestUuid {

    @Id
    @Column(name = "id_uuid")
    @GeneratedValue(strategy = GenerationType.AUTO)
    private UUID idUuid;

    @Column(name = "name")
    private String name;

    /**
     * @return the idUuid
     */
    public UUID getIdUuid() {
        return idUuid;
    }

    /**
     * @param idUuid the idUuid to set
     */
    public void setIdUuid(UUID idUuid) {
        this.idUuid = idUuid;
    }

    /**
     * @return the name
     */
    public String getName() {
        return name;
    }

    /**
     * @param name the name to set
     */
    public void setName(String name) {
        this.name = name;
    }

}

When I retrieve the table values I don't get the same value for ID_UUID whats stored in the database because of UUID data type (if I change it to a String I am able to pull the correct value). My limitation is to use the UUID type as I am migrating from Postgre to Oracle database and cannot change my entity class. Is it possible to retrieve the correct value using UUID data type?

CodePudding user response：

You need to define custom type that stores and retrieves UUID type as VARCHAR.

See Hibernate UUID type implementation.
https://github.com/hibernate/hibernate-orm/blob/main/hibernate-core/src/main/java/org/hibernate/type/descriptor/java/UUIDJavaType.java
In case if you have Hibernate 5 or higher you can use it.

1. Create own user type

public class UUIDType extends AbstractSingleColumnStandardBasicType<UUID> {

    public static final UUIDType INSTANCE = new UUIDType();
    public static final String NAME = "UUIDType";

    public UUIDType() {
        super(VarcharTypeDescriptor.INSTANCE, UUIDTypeDescriptor.INSTANCE);
    }

    @Override
    public String getName() {
        return NAME;
    }
}

2. Implement a UUIDTypeDescriptor for storing and retrieving UUID as VARCHAR

public class UUIDTypeDescriptor extends AbstractTypeDescriptor<UUID> {

    public static final UUIDTypeDescriptor INSTANCE = new UUIDTypeDescriptor();

    public UUIDTypeDescriptor() {
        super(UUID.class, ImmutableMutabilityPlan.INSTANCE);
    }

    @Override
    public String toString(UUID uuid) {
        return uuid.toString();
    }

    @Override
    public UUID fromString(String s) {
        return UUID.fromString(s);
    }

    @Override
    public <T> T unwrap(UUID uuid, Class<T> type, WrapperOptions wrapperOptions) {
        if (uuid == null) return null;

        if (String.class.isAssignableFrom(type)) {
            return (T) uuid.toString();
        }

        throw unknownUnwrap(type);
    }

    @Override
    public <T> UUID wrap(T value, WrapperOptions wrapperOptions) {
        if (value == null)
            return null;

        if(value instanceof String) {
            return UUID.fromString((String) value);
        }

        throw unknownWrap(value.getClass());
    }
}

3. Apply custom type to your entity

Specific for entity:

@TypeDef(name = UUIDType.NAME, 
         typeClass = UUIDType.class,
         defaultForType = UUID.class)
@Entity
@Table(name = "test_uuid")
public class TestUUID {

}

or on package lavel in package-info.java file:

@TypeDef(
    name = UUIDType.NAME,
    typeClass = UUIDType.class,
    defaultForType = UUID.class
)
package com.model;