How to map a TIMESTAMP column to a ZonedDateTime JPA entity property?

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I'm using Spring data jpa and mariadb latest version, and MariaDB 10.3.16

+--- org.springframework.boot:spring-boot-starter-data-jpa -> 2.1.5.RELEASE
...
|    +--- org.springframework.boot:spring-boot-starter-jdbc:2.1.5.RELEASE
...
|    +--- org.hibernate:hibernate-core:5.3.10.Final

This is my Entity:

@Entity
@Data
@Table
@NoArgsConstructor
@AllArgsConstructor
public class Note {
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Integer id;

    @Column
    private String gsn;

    @Column
    @Enumerated(EnumType.STRING)
    private NoteType type;

    @Column
    private String text;

    @Column
    private ZonedDateTime scheduleDt;

    @Column
    @CreationTimestamp
    private Instant createDt;

    @Column
    @UpdateTimestamp
    private ZonedDateTime updateDt;
}

When I persist my entity, Hibernate tries to save ZonedDateTime member as DATETIME column. But I want to use TIMESTAMP column instead of DATETIME column.

This is create DDL, what I see from log.

create table `note` (`id` integer not null, `create_dt` datetime,
    `gsn` varchar(255), `schedule_dt` datetime, `text` varchar(255),
    `type` varchar(255), `update_dt` datetime, primary key (`id`)) 
  engine=MyISAM

Here create_dt, schedule_dt, update_dt is created as datetime column type, what is not I wanted. (I don't like MyISAM, too).

How can I fix it?

Added because comment cannot express ddl.

When I use columnDefinition attribute, generated ddl is ...

create table `note` (`id` integer not null, `create_dt` datetime,
    `gsn` varchar(255), `schedule_dt` datetime, `text` varchar(255),
    `type` varchar(255), `update_dt` `TIMESTAMP`, primary key (`id`))

engine=MyISAM

There is unrequired '`' around TIMESTAMP.

zoneddatetime mariadb jpa hibernate java
2 Answers

JPA 2.2 offers support for mapping Java 8 Date/Time API, but only for the following types:

However, Hibernate supports also ZonedDateTime, like this.

When saving the ZonedDateTime, the following Timestamp is going to be sent to the PreparedStatement:

Timestamp.from( zonedDateTime.toInstant() )

And, when reading it from the database, the ResultSet will contain a Timestamp that will be transformed to a ZonedDateTime, like this:

ts.toInstant().atZone( ZoneId.systemDefault() )

Note that the ZoneId is not stored in the database, so basically, you are probably better off using a LocalDateTime if this Timestamp conversion is not suitable for you.

So, let's assume we have the following entity:

@Entity(name = "UserAccount")
@Table(name = "user_account")
public class UserAccount {

    @Id
    private Long id;

    @Column(name = "first_name", length = 50)
    private String firstName;

    @Column(name = "last_name", length = 50)
    private String lastName;

    @Column(name = "subscribed_on")
    private ZonedDateTime subscribedOn;

    //Getters and setters omitted for brevity
}

Notice that the subscribedOn attribute is a ZonedDateTime Java object.

When persisting the UserAccount:

UserAccount user = new UserAccount()
    .setId(1L)
    .setFirstName("Vlad")
    .setLastName("Mihalcea")
    .setSubscribedOn(
        LocalDateTime.of(
            2020, 5, 1,
            12, 30, 0
        ).atZone(ZoneId.systemDefault())
    );

entityManager.persist(user);

Hibernate generates the proper SQL INSERT statement:

INSERT INTO user_account (
    first_name, 
    last_name, 
    subscribed_on, 
    id
) 
VALUES (
    'Vlad', 
    'Mihalcea', 
    '2020-05-01 12:30:00.0', 
    1
)

When fetching the UserAccount entity, we can see that the ZonedDateTime is properly fetched from the database:

UserAccount userAccount = entityManager.find(
    UserAccount.class, 1L
);

assertEquals(
    LocalDateTime.of(
        2020, 5, 1,
        12, 30, 0
    ).atZone(ZoneId.systemDefault()),
    userAccount.getSubscribedOn()
);
user_10251180

You can define the column type using the @Column annotation:

@Column(columnDefinition="TIMESTAMP")  
@UpdateTimestamp
private ZonedDateTime updateDt;
user_10451420
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